Chemistry NCERT Exemplar Solutions Class 11th Chapter One: Overview, Questions, Preparation

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According to the law of multiple proportions, when two elements react to form two or more than two chemical compounds, the ratio between different masses of one of the elements combining with a fixed mass of the other is always in the ratio of tiny numbers.

 Example:

1. Compounds of carbon and oxygen:

C and O react to form two different compounds CO and CO₂. In CO, 12 parts by mass of C reacts with 16 parts by mass of 0 .

In CO₂ ,12 parts by mass of C reacts with 32 parts by mass of O .

If the mass of C is fixed at 12 parts of mass then the ratio in the masses of oxygen which reacts with the fixed mass of C is 16: 32, that is, 1: 2 .

Therefore, the mass of oxygen contains a simple ratio of 1: 2 to each other.

2. Compounds of sulphur and oxygen:

S and O reacts to form two compounds SO₂ and SO_{3 .}

 In SO₂ ,32  parts by mass of S reacts with 32 parts by mass of O . In SO₃ ,32  parts by mass of S reacts with 48 parts by mass of O .

If the mass of S is fixed at 32 parts of mass then the ratio in the masses of oxygen which react with the fixed mass of S is 32: 48, that is, 2: 3.

This law demonstrates that certain constituents combine in a specific proportion. It's possible that these constituents are atoms. As a result, the law of multiple proportions tells the existence of atoms that can join to form molecules.

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P₁=1 atm

          P₂= 1/2=0.5 atm

          T₁=273.15 K

          V₂=?

          V₁=?

32 g dioxygen occupies = 22.4 L volume at STP

∴ 1.6 g dioxygen will occupy = 22.4L x 1.6g / 32g = 1.12 L

     V₁=1.12 L

From Boyle's law (as temperature is constant)

p₁V₁=p₂V₂

V₂=p₁V₁p₂

    = 1 atm x 1.12 l/0.6 atm g = 2.24 L

(ii) Number of molecules of dioxygen.

        Ans-  (ii) Number of moles of dioxygen= Mass of dioxygen / Molar mass of dioxygen

                                              nO₂ = 1.6/32 = 0.05 mol
                  1 mol of dioxygen contains = 6.022×10²³ molecules of dioxygen
                  ∴ 0.05 mol of dioxygen = 6.022×10²³×0.05 molecule of O₂

                = 0.3011×10²³ molecules

                =3.011×10²² molecules

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The volume of HCl solution is 250 mL and its molarity is 0.76M.

The number of moles of HCl as follows,

Moles of HCl Molarity Volume (in L)

= 0.76M 0.250 L

= 0.19 mol

The molar mass of CaCO₃ is 100 g / gQl and the mass of CaCO₃ is given as 1000 g

The number of moles of CaCO_{3 i}s calculated as

Moles of CaCO₃ =     M a s  / M o l a r   m a s       

=      1000 g / 100 g   /   m o l  = 10 mol

According to the given reaction, 1 mole of CaCO₃ requires 2 moles of HCl. So, the required number of moles of HCl for 10 moles of CaCO₃ is calculated as

Moles of HCl =   2mol of HCl /1mol of CaCO₂   * 10 mol of CaCO₃

=  20 mol

So, the required number of moles of HCl is 20 mol but only 0.19 mol are given. So, HCl is a limiting reagent. So, the amount of calcium chloride formed is depend upon the limiting reagent, that is, the amount of HCl available.

According to the reaction, 2 moles of HCl gives 1 mol of CaCl₂ Sg 2 ., the number of moles of calcium chloride produced by 0.19 mol of HCl as follows,

Moles of CaCl₂ =  1 mol of CaCl₂ / 2 mol  of HCl  x 0.19 mol of HCl

                           = 0.095 mol

The molar mass of calcium chloride is 111 g / mol. So, its mass is calculated as

Mass of CaCl₂ = Moles x  Molar mass

= 0.095 mol x 111 g / mol

= 10.54 g

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In AB, 2 g of A combines with 5 g of B.

So, 4 g of A combines with 10 g of B.

In AB₂, 2g of B combines with 10 g of B, So 4g of A combines with 20 g of B.

In A₂B₃ , 4G of B combines with 5 g of B.

In A₂B₃ 4 g of B combined with 15 g of B.

So, the ratio between different masses of B which combine with fixed mass (4g) of A is 10: 20: 5: 15, that is, 2: 4: 1: 3.

Hence, the ratio is simple. Therefore, the law of multiple proportions is applicable.

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Mass of 1 mole of C-12 = 12g

1 mole contains 6.022×10²³ atoms.

Thus, mass of 6.022×10²³ atoms=12g

 Mass of 1 atom of carbon =126.022×10²³ g
                                         =1.99×10⁻²³ g

Thus, mass of one atom of C-12 is 1.99×10⁻²³ g

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On solving the above equation, the result is 5.4. All non-zeroes digits are significant. The significant figure is 2.

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SI unit of the mole is mol. The amount of a substance that contains as many particles or entities as there are atoms in exactly

12 g (0.012 kg) of the C-12  isotope is defined as a mole. One mole is defined as follows:

1 mole =  6.023 x 10²³

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The number of moles of a substance (known as the solute) dissolved in precisely 1 litce of a solution is known as molarity (solvent and solute combined). As a result, the formula for estimating molarity is as follows:

Molarity = $\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{e}}{\mathrm{v}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{}\mathrm{i}\mathrm{n}\mathrm{}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{r}\mathrm{e}\mathrm{}}$

The term molarity is also used to refer to molar concentration. As a result, molar concentration measurement is based on the volume of liquid in which a substance is dissolved. It's vital to remember that the volume is in litres, so if we have volume in mL we need to convert that in liters.

Molality is the number of moles of substance (also known as the solute) found in a given mass of solvent (in Kg ) in which it is dissolved.

 Molality is calculated by using the formula.

Major differences between Molarity and Molality are given as follows:

 1) Molarity is the concentration of a material determined as the number of moles of solute dissolved in poe litre of solution, whereas molality is the concentration calculated as the number of moles of solute found in one kilogram of solvent.

2) Molality is denoted by the symbol m, whereas molarity is denoted by the symbol M.

3) The molarity formula is moles per litre, but the molality formula is moles per kilogram.

4) Molarity is impacted by temperature changes, whereas molality is unaffected by temperature changes.

5) Changes in pressure affect molarity, but they do not affect molality.

6) Molarity can lead to an imprecise and inaccurate concentration, whereas molality can lead to an accurate and precise concentration measurement.

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Molecular mass of Ca₃ (PO₄)₂ = 310.18 g / mole 

Given mass of calcium 4 x 30= 120 g

Given mass of phosphorous =  31 x 2 =62 g

Given mass of oxygen = 16 x 8 = 128 g

Mass percent of Calcium = $\frac{120}{310.18\mathrm{}}$ x 100 = 38.71%

Mass percent of Phosphorous = $\frac{62}{310.18\mathrm{}}$  x 100 = 20%

Mass percent of Oxygen = $\frac{128}{310.18\mathrm{}}$ x 100 = 41.29 %

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If all gases are at the same temperature and pressure, Gay lussac's law of gaseous volumes states that gases combine or are created in a chemical reaction in a simple volume ratio.

H₂ (g) +          Cl₂ (g) →           2HCL (g)

1 volume      1 volume        2 volume

22.4 litre       22.4 litre        44.8 litre

2N₂ (g) +      O₂ (g) →            2N₂O (g)

2Vol             1Vol                  2Vol

45.4 litre     22.7 litre          45.4 litre

2: 1: 2

The above examples proved that gaseous volumes combine in a simple volume ratio.

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(a) Is this statement true?  

Ans: Yes the given statement is true

(b) If yes, according to which law?

Ans: Multiple law of proportions: According to the Law of Multiple proportions, when two elements combine to generate more than one compound, the weights of one element that combine with a fixed weight of the other are in a ratio of tiny whole numbers.

(c) Give one example related to this law

Ans:

C (g) + O (g) -> CO (g)

12 g    16 g       28 g

C (g) + O₂ (g) -> CO₂ (g)

12 g     32 g       44 g

This mass of Oxygen, which combines with a fixed mass of Carbon is in the simple ratio that is 16:32 or 1:2.

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(Natural abundance of 1H x molar mass ) + (Natural abundance of ²H x molar mass of 2H)

Natural abundance of  ¹H = 99.985

Natural abundance of ²H = 0.015

Average atomic mass = $\frac{99.985+0.030}{100\mathrm{}}$

$\frac{100.015}{100\mathrm{}}$ = 1.00015u

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65.3 g of Zinc gives 22.7 litres of Hydrogen gas

32.65 g Zinc gives = 32.65g x 22.7 litres/65.3 = 11.35 L

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Mass of NaoH = 40 g

Mass of solvent =1000 g

Mass of solution = 40 x 3+1000

Density = $\frac{\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{}}{\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}}$

$\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}$ = $\frac{\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{}}{\mathrm{D}\mathrm{e}\mathrm{n}\mathrm{s}\mathrm{i}\mathrm{t}\mathrm{y}}$

$\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}$ = $\frac{1120\mathrm{g}}{1.10\mathrm{g}/\mathrm{m}\mathrm{L}}$= 1009.0 mL

Molarity = $\frac{\mathrm{N}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{e}\mathrm{}}{\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{i}\mathrm{t}\mathrm{y}}$  = $\frac{3}{1.009\mathrm{}}=$  2.97M

1009.00 mL= 1.009 L

Hence, the molarity of the solution is 2.97M

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Molality is the number of moles of substance (also known as the solute) found in a given mass of solvent (in Kg ) in which it is dissolved. Molality is calculated by using the formula.

Molality = $\frac{\mathrm{N}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{e}\mathrm{}}{\mathrm{S}\mathrm{o}\mathrm{l}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{}\mathrm{i}\mathrm{n}\mathrm{}\mathrm{k}\mathrm{g}\mathrm{}}$

So, temperature has no effect on the molality of the solution because molality is expressed in mass.

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Molar mass of NaOH = 40 g/ mole

Molar mass of water =18 g / mole

Mass of NaOH= 4 g

Number of moles of 4 g NaOH = $\frac{4\mathrm{}\mathrm{g}}{40\mathrm{}\mathrm{g}\mathrm{}}$ = 0.1 mol

Number of moles of H₂O= $\frac{36\mathrm{}\mathrm{g}}{18\mathrm{}\mathrm{g}\mathrm{}}$ =2 mol

Mole fraction of water

$\frac{26\mathrm{}\mathrm{g}}{\mathrm{N}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}+\mathrm{}\mathrm{N}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}{\mathrm{H}}_2\mathrm{O}}$

Mole fraction of NaOH = $\frac{2\mathrm{}\mathrm{g}}{0.1\mathrm{}+\mathrm{}2\mathrm{}\mathrm{g}\mathrm{}}$  = $\frac{2\mathrm{}\mathrm{g}}{2.1\mathrm{}\mathrm{g}\mathrm{}}$  = 0.95

Number of moles of= $\frac{2\mathrm{}\mathrm{g}}{0.1\mathrm{}\mathrm{g}\mathrm{}}$  = $\frac{0.1\mathrm{}\mathrm{g}}{2.1\mathrm{}\mathrm{g}\mathrm{}}$ = 0.047

Number of moles of NaOH

Mass of solution = Mass of solute + Mass of solvent

Mass of NaOH + Mass of water 4 g + 36 g = 40 g

Specific gravity of solution =1 g / ml

1 litre =1000 ml volume of solution = 40ml

40ml = 0.04 litre

Molarity = $\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{e}}{\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{}\mathrm{i}\mathrm{n}\mathrm{}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{r}\mathrm{e}}$ = $\frac{0.1\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}}{0.04\mathrm{}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{r}\mathrm{e}\mathrm{}}$ = 2.5M

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The reaction is shown below.

2A + 4B? 3C + 4D 

According to the above equation, 2 moles of A requires 4 moles of B. So, the number of moles of B required for 5 moles of A is calculated as,

Moles of B = 5 mol of A * $\frac{4\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{B}}{2\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{A}\mathrm{}}$ = 10 mol of B

So, the required number of moles of B is 10 mal but only 6 moles of B are given in the question. Therefore, B is the limiting reagent.

(ii) calculate the amount of C formed?

Ans: Now, the amount of C can be calculated by the limiting reagent, that is, the amount of B.

According to the equation, 4 moles of B gives 3 moles of C . So, the number of moles of C formed by 6 moles of B as follows,

                      Moles of C = 6 mol of B * $\frac{3\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{C}}{4\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{B}\mathrm{}}$ = 4.5 mol of C

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Option (ii)

Average of readings of student A,

Average of student A = $\frac{3.01+2.99}{2}=3.00$

Average of readings of student B,

Average of student B = $\frac{3.05+2.95}{2}=3.00$

The correct reading is 3. For both A and B, the average value is close to the correct value. Thus, readings of both are accurate.

The readings of student A are also very close to each other and close to the average value. Thus, readings are precise. Also, the readings of student B are also very close to each other and close to the average value.

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Option (iii)

For the conversion from Fahrenheit to Celsius

$\mathrm{F}=\frac{9}{5}\mathrm{C}+32$

 On substituting the values in the above equation,

 200 = $\frac{9}{5}C+32$

         =93.3 $°$  C

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Option (C)

No. of mole given = $\frac{\mathrm{g}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{}}{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}}$

On substituting the value in the above equation, the cal, can be calculated as

no, of mole =  $\frac{5.85\mathrm{}\mathrm{g}\mathrm{}}{58.5\mathrm{}\mathrm{g}\mathrm{}{\mathrm{m}\mathrm{o}\mathrm{l}}^{-1}}$  = 0.1 g

The molarity (M) is given by the formula:

M = $\frac{\mathrm{n}}{{\mathrm{V}}_1}$

On substituting the values in the above equation:

Molarity = $\frac{0.1\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{}1000\mathrm{}\mathrm{L}}{500}$

= 0.2 mol L^-1

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Option (B)

The relation between molarity and volume is given as,

M₁ V₁= M₂ V₂

On substituting the value in the above equation, the political can be calculated as

5M* 500 mL = M₂*1500 mL 

M =1.66M

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Option (D)

(A) The number of moles is given by the following formula,

Moles = $\frac{\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}}{\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{}}\dots$ ….(1)

The number of moles of He is calculated by using equation (1) as follows

Moles of O₂ = $\frac{4\mathrm{}\mathrm{g}}{4\mathrm{}\mathrm{g}\mathrm{}/\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{}}$  =1 mol

The number of atoms can be calculated as, number of moles

$\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{m}\mathrm{s}}{\mathrm{A}\mathrm{v}\mathrm{o}\mathrm{g}\mathrm{a}\mathrm{d}\mathrm{r}\mathrm{o}\mathrm{}\mathrm{s}\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}}\dots$ ….(2)

On substituting the values in the above equation:

1 mol = $\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{m}\mathrm{s}\mathrm{}}{6.022\times \mathrm{}{10}^{23}\mathrm{}}$

Number of atoms = 1 $\times$ 6.022 $\times$ 10²³

(B) The number of moles of Na is calculated by using equation (1) as follows,

Moles of Na = $\frac{46\mathrm{}\mathrm{g}\mathrm{}}{23\mathrm{}\mathrm{g}\mathrm{}/\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{}\mathrm{}}$  = 2 mol

The number of atoms can be calculated by using equation (2) as follows,

2 mol = $\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{m}\mathrm{s}}{6.022\times \mathrm{}{10}^{23}}$

number of atoms= 2 $\times$  6.022 $\times$ 10²³

(C) The number of moles of Ca is calculated by using equation (1) as follows

Moles of Ca= $\frac{0.40\mathrm{}\mathrm{g}}{40\mathrm{}\mathrm{g}\mathrm{}/\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{}\mathrm{}}$  = 0.01 mol

The number of atoms can be calculated by using equation (2) as follow

0.01 mol = $\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{m}\mathrm{s}}{6.022\times \mathrm{}{10}^{23}}$

number of atoms = 0.01 $\times$  6.022 $\times$ 10²³

(D) The number of moles of He is calculated by using equation (1) as follows,

Moles of He = $\frac{12\mathrm{}\mathrm{g}}{4\mathrm{}\mathrm{g}\mathrm{}/\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{}\mathrm{}\mathrm{}}$  = 3 mol

The number of atoms can be calculated by using equation (2) as follows,

3 mol= $\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{m}\mathrm{s}}{6.022\times \mathrm{}{10}^{23}}$

number of atoms = 3 $\times$  6.022 $\times$ 10²³

Thus, 12 g He contains the highest number of atoms.

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Option (C)

The molar mass of glucose is  180 g mol^- . The molarity (M) is given by the formula:

M = $\frac{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}}{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}}$

On substituting the value in the above equation, the mol can be calculated as

M  = $\frac{0.90\mathrm{}\mathrm{g}\mathrm{}{\mathrm{L}}^{-1}}{180\mathrm{}\mathrm{g}\mathrm{}{\mathrm{m}\mathrm{o}\mathrm{l}}^{-1}}$ = 0.005M

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Option (D)

The number of moles is given by the following formula,

Moles = $\frac{\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}}{\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}}$  …. (1)

The number of moles of HCl is calculated by using equation (1) as follows,

Moles of HCl $\mathrm{}=\frac{18.25\mathrm{}\mathrm{g}}{36.5\mathrm{}\mathrm{g}\mathrm{}/\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}}$ 0.5 mol

The molality (m) is given by the formula:

m = $\frac{\mathrm{n}}{\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}{\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{t}}^{\mathrm{k}\mathrm{g}}}$

On substituting the values in the above equation:

Molality = $\frac{0.5\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{}1000\mathrm{}\mathrm{k}\mathrm{g}}{500}$ = 1m

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Option (A)

The molarity (M) is given by the formula:

M $=$ $\frac{\mathrm{n}}{{\mathrm{V}}_{\left(\mathrm{L}\right)}\mathrm{}}$

On substituting the values in the above equation:

0.02M = $\frac{\mathrm{n}\times \mathbf{}1000\mathrm{L}}{100\mathrm{}}$

n = 0.002 mol

The number of molecules can be calculated as, number of moles

= $\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{s}}{\mathrm{A}\mathrm{v}\mathrm{o}\mathrm{g}\mathrm{a}\mathrm{d}\mathrm{r}\mathrm{o}\mathrm{}\mathrm{s}\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}}$

On substituting the values in the above equation:

0.002 mol = $\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{s}}{6.022\times \mathrm{}{10}^{23}\mathrm{}}$

number of molecules = 0.002 $\times$ 6.022 $\times$ 10²³

= 12.044 $\times$  10²⁰

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Option (B)

The molar mass of carbon is 44 g mol^-1

44 g of carbon dioxide contains 12 g of carbon

The percentage composition is given as, % composition

= $\frac{\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{a}\mathrm{t}\mathrm{}\mathrm{c}\mathrm{l}\mathrm{e}\mathrm{m}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{}}{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{o}\mathrm{u}\mathrm{n}\mathrm{d}\mathrm{}\mathrm{}}$ $*100$

On substituting the values in the above equation,

% of carbon = $\frac{12\mathrm{}\mathrm{g}}{44\mathrm{}\mathrm{g}\mathrm{}{\mathrm{m}\mathrm{o}\mathrm{l}}^{-1}\mathrm{}\mathrm{}}$ $*100$

= 27.27%

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Option (C)

The empirical formula mass of CH₂O is 30 g

The relation between empirical and molecular formula is given as,

n = $\frac{\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{u}\mathrm{l}\mathrm{a}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}}{\mathrm{E}\mathrm{m}\mathrm{p}\mathrm{i}\mathrm{r}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{u}\mathrm{l}\mathrm{a}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{}\mathrm{}\mathrm{}}$  …. (1)

On substituting the values in equation (1)

n = $\frac{180\mathrm{}\mathrm{g}}{30\mathrm{}\mathrm{g}\mathrm{}\mathrm{}}$  = 6

Thus, the molecular formula of the compound will be,

Molecular formula of the compound = (CH₂O)₆ = C₆H₁₂O₆

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Option (A)

The density (d ) is given by the formula:

d = $\frac{\mathrm{m}}{\mathrm{V}\mathrm{}\mathrm{}}$

Here, m is the mass and v is the volume

On substituting the values in the above equation:

3.12 g mL^-1 = $\frac{\mathrm{m}}{1.5\mathrm{}\mathrm{m}\mathrm{L}\mathrm{}\mathrm{}}$ M = 4.68 g

The mass will be 4.7 upto 2 significant figure.

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Option (C)

(A) The molecule of a compound is obtained when two or more atoms of different elements are combined together.

(B) A compound can be separated into its constituent elements by physical methods of separation.

(C) A compound does not retains the physical and chemical properties of its constituent elements.

(D) When two elements react to form two or more than two chemical compounds, the ratio between different masses of one of the elements combining with a fixed mass of the other.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (A)

According to the law of conservation of mass, matter can neither be created nor be destroyed. In the given reaction,

4Fe + 3O₂→ 2Fe₂O₃

The mass of reactant is equal to the mass of product. Hence, it follows the law of conservation of mass

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (B)

In a, c, and d options given reactions, all the elements are balanced and hence, following the law of conservation of mass.

In reaction (B), C₃H₈+ O₂→ CO₂+H₂0  elements are not balanced. Hence product are different. It is against the law of conservation of mass

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (B)

Law of multiple proportion: According to the law of multiple proportion, whe form two or more than two chemical compounds, the ratio between different elements combining with a fixed mass of the other is always in the ratio of tin Compounds of carbon and oxygen:

Carbon and oxygen react to form two different compounds CO and CO2. In Carbon react with 16 parts by mass of oxygen.

 In CO₂ ,12  parts by mass of Carbon react with 32 parts by mass of oxygen. If the mass of Carbon is fixed at 12 parts of mass then the ratio in the masses with the fixed mass of carbon is 16: 32, that is, 1: 2. Therefore, the mass of oxygen contains a simple ratio of 2: 1.to each other

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (A), (D)

One mole of oxygen gas at STP is equal to  6.022 x 10²³  molecules of oxygen The number of moles is given by the following formula,

Moles = $\frac{\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}}{\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{}\mathrm{}}$    ----------(1)

The number of moles of O₂ is calculated by using equation (1) as follow.

Moles of O₂  = $\frac{16\mathrm{}\mathrm{g}}{32\mathrm{}\mathrm{g}\mathrm{}/\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{}\mathrm{}}$

                       = 0.5 mol

The number of molecules can be calculated as, number of moles

= $\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{s}}{\mathrm{A}\mathrm{v}\mathrm{o}\mathrm{g}\mathrm{a}\mathrm{d}\mathrm{r}\mathrm{o}\mathrm{}\mathrm{s}\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}}$   ----------(2)

On substituting the values in the above equation:

0.5 mol = $\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{s}}{6.022\times \mathrm{}{10}^{23}\mathrm{}\mathrm{}}$

number of molecules = 0.5 $\times$  6.022 $\times$ 10²³

The number of moles is given by the following formula,

moles = $\frac{\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}}{\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{}\mathrm{}}$  ----------(1)

The number of moles of O₂ is calculated by using equation (1) as follows,

Moles of O₂ = $\frac{32\mathrm{}\mathrm{g}}{32\mathrm{}\mathrm{g}\mathrm{}/\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}}$

=1 mol

The number of molecules can be calculated as, number of moles

= $\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{s}}{\mathrm{A}\mathrm{v}\mathrm{o}\mathrm{g}\mathrm{a}\mathrm{d}\mathrm{r}\mathrm{o}\mathrm{}\mathrm{s}\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}}$ ----------(2)

On substituting the values in the above equation:

1 mol = $\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{s}}{6.022\times \mathrm{}{10}^{23}\mathrm{}\mathrm{}}$

Number of molecules = 1 $\times$  6.022 $\times$ 10²³

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (B), (C)

0.1 mole H₂SO₄ reacts with 1 mole of NaOH.

0.1 mole of NaOH will react with $\frac{0.1}{2\mathrm{}\mathrm{}}$ = mole of H₂SO₄

Here, NaOH is the limiting reagent.

2 mole of NaOH produces 1 mole of Na₂SO₄

0.1 mole of NaOH will give $\frac{0.1}{2\mathrm{}\mathrm{}}$  mole of Na₂SO₄

No. of mole = $\frac{\mathrm{g}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}}{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{}\left({\mathrm{N}\mathrm{a}}_2{\mathrm{S}\mathrm{O}}_4\right)\mathrm{}\mathrm{}\mathrm{}}$

On substituting the value in the above equation, the mass can be calculated as

0.05 mol = $\frac{\mathrm{g}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}}{142\mathrm{}{\mathrm{g}\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}-}^1\mathrm{}\mathrm{}\mathrm{}\mathrm{}}$

given mass = 7.10 g

Volume of solution after mixing is 2 L.

So, the molarity of Na₂SO₄ is

Molarity = $\frac{0.05}{2\mathrm{}\mathrm{}\mathrm{}\mathrm{}}$  = 0.025mol L^-1

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option(C), (D)

(A) The number of moles is given by the following formula,

Moles = $\frac{\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}}{\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}}\mathrm{}$ -----------(1)

The number of moles of O₂ is calculated by using equation (1) as follows,

Moles of O₂ = $\frac{16\mathrm{}\mathrm{g}}{32\mathrm{}\mathrm{g}\mathrm{}/\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}}$  = 0.5 mol

The number of atoms can be calculated as, number of moles

$\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{m}\mathrm{s}\mathrm{}}{\mathrm{A}\mathrm{v}\mathrm{o}\mathrm{g}\mathrm{a}\mathrm{d}\mathrm{r}\mathrm{o}\mathrm{}\mathrm{s}\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{}}\mathrm{}$ -----------(2)

On substituting the values in the equation (2)

0.5 mol = $\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{m}\mathrm{s}\mathrm{}}{6.022\times \mathrm{}{10}^{23}\mathrm{}\mathrm{}\mathrm{}\mathrm{}}$

number of atoms = 0.5 $\times \mathrm{}$ 6.022 $\times \mathrm{}$ 10²³

The number of moles of H₂ is calculated by using equation (1) as follows,

Moles of H₂ = $\frac{4\mathrm{g}}{2\mathrm{}\mathrm{g}\mathrm{}/\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}}$

= 2 mol

On substituting the values in the equation (2):

2 mol = $\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{m}\mathrm{s}\mathrm{}}{6.022\times \mathrm{}{10}^{23}\mathrm{}\mathrm{}\mathrm{}\mathrm{}}$

number of atoms = 2 $\times \mathrm{}$ 6.022 $\times \mathrm{}$ 10²³

(B) The number of moles of CO₂ is calculated by using equation (1) as follows

Moles of CO₂ = $\frac{44\mathrm{}\mathrm{g}}{44\mathrm{}\mathrm{g}\mathrm{}/\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}}$

= 1 mol

On substituting the values in the equation (2), the number of atoms can be calculated as ,

1 mol = $\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{m}\mathrm{s}\mathrm{}}{6.022\times \mathrm{}{10}^{23}\mathrm{}\mathrm{}\mathrm{}\mathrm{}}$

number of atoms = 1 $\times \mathrm{}$ 6.022 $\times \mathrm{}$ 10²³

(C) The number of moles of N₂ is calculated by using equation (1) as follows,

Moles of N₂ = $\frac{28\mathrm{g}}{28\mathrm{}\mathrm{g}\mathrm{}/\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{}\mathrm{}\mathrm{}\mathrm{}}$  =1 mol

On substituting the values in the equation (2), the number of molecules can be calculated as

1 mol = $\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{m}\mathrm{s}\mathrm{}}{6.022\times \mathrm{}{10}^{23}\mathrm{}\mathrm{}\mathrm{}\mathrm{}}$

number of atoms = 1 $\times \mathrm{}$ 6.022 $\times \mathrm{}$ 10²³

The number of moles of O₂ is calculated by using equation (1) as follows,

Moles of O₂ = $\frac{32\mathrm{}\mathrm{g}}{32\mathrm{}\mathrm{g}\mathrm{}/\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}}$  = 1

The number of atoms can be calculated as, number of moles

$\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{m}\mathrm{s}\mathrm{}}{\mathrm{A}\mathrm{v}\mathrm{o}\mathrm{g}\mathrm{a}\mathrm{d}\mathrm{r}\mathrm{o}\mathrm{}\mathrm{s}\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{}}\mathrm{}$ -

On substituting the values in the equation (2)

1 mol = $\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{m}\mathrm{s}\mathrm{}}{6.022\times \mathrm{}{10}^{23}\mathrm{}\mathrm{}\mathrm{}\mathrm{}}$

number of atoms = 1 $\times \mathrm{}$ 6.022 $\times \mathrm{}$ 10²³

(D) The number of moles of C is calculated by using equation (1) as follows,

Moles of C = $\frac{12\mathrm{}\mathrm{g}\mathrm{}}{12\mathrm{}\mathrm{g}\mathrm{}/\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{}\mathrm{}}$ =1 mol

On substituting the values in the equation (2), the number of molecules can be calculated as

1 mol = $\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{m}\mathrm{s}\mathrm{}}{6.022\times \mathrm{}{10}^{23}\mathrm{}\mathrm{}\mathrm{}\mathrm{}}$

number of atoms = 1 $\times \mathrm{}$ 6.022 $\times \mathrm{}$ 10²³

The number of moles of Na is calculated by using equation (1) as fo

Moles of Na = $\frac{23\mathrm{}\mathrm{g}\mathrm{}}{23\mathrm{}\mathrm{g}\mathrm{}/\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{}\mathrm{}}$ =1 mol

On substituting the values in the equation (2), the number of molecules can be calculated as

1 mol = $\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{m}\mathrm{s}\mathrm{}}{6.022\times \mathrm{}{10}^{23}\mathrm{}\mathrm{}\mathrm{}\mathrm{}}$

number of atoms = 1 $\times \mathrm{}$ 6.022 $\times \mathrm{}$ 10²³

This is a Multiple Choice Questions as classified in NCERT Exemplar

The answers are options (i) and (ii)

(A)The number of moles is given by the following formula,

Moles = $\frac{\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}}{\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}}$

So, the number of moles of NaOH is calculated, by using equation (1) as follows

 20 g NaOH in 200 mL solution.

Moles of NaOH= $\frac{20\mathrm{g}}{40\mathrm{g}/\mathrm{m}\mathrm{o}\mathrm{l}}=0.5$   mol of NaOH

The molarity (M ) is given by the formula:

M= $\frac{\mathrm{n}}{{\mathrm{V}}_{\mathrm{L}}\mathrm{}\mathrm{}}$

On substituting the values in the above equation:

M _(NaOH) = $\frac{0.5\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{}*\mathrm{}1000}{200\mathrm{}\mathrm{L}\mathrm{}\mathrm{}}$

= 2.5 mol L^-1

(B) The molarity is given by the formula:

M= $\frac{\mathrm{n}}{{\mathrm{V}}_{\mathrm{L}}\mathrm{}\mathrm{}}$

On substituting the values in the above equation:

M _(KCL) = $\frac{0.5\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{}*1000}{200\mathrm{}\mathrm{L}\mathrm{}\mathrm{}}$

= 2.5 mol L^-1

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option(C), (D)

The number of moles is given by the following formula,

Moles = $\frac{\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}}{\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}}$  ---------------(1)

The number of molecules can be calculated as, number of moles

= $\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{s}}{\mathrm{A}\mathrm{v}\mathrm{o}\mathrm{g}\mathrm{a}\mathrm{d}\mathrm{r}\mathrm{o}\mathrm{}\mathrm{s}\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}}$  ----------(2)

(C) The number of moles of N₂ is calculated by using equation (1) as follows,

Moles of N₂ = $\frac{14\mathrm{}\mathrm{g}}{28\mathrm{}\mathrm{g}\mathrm{}/\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{}}$  = 0.5 mol

The number of molecules can be calculated by using equation (2) as follows,

0.5 mol = $\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{s}}{6.022\times \mathrm{}{10}^{23}\mathrm{}\mathrm{}}$

number of molecules = 0.5 $\times$  6.022 $\times$  10²³

(D) The number of moles of H₂ is calculated by using equation (1) as follows,

Moles of H₂ = $\frac{1\mathrm{g}}{2\mathrm{}\mathrm{g}\mathrm{}/\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{}}$  = 0.5 mol

The number of molecules can be calculated by using equation (2) as follows,

0.5 mol = $\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{s}}{6.022\times \mathrm{}{10}^{23}\mathrm{}\mathrm{}}$

number of molecules = 0.5 $\times$ 6.022 $\times$  10²³

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (C), (D)

Molality: It is defined as the number of moles of solute divided by the mass of solvent in kg . It is denoted by m . The formula is expressed as

m = $\frac{\mathrm{n}}{\mathrm{W}\mathrm{}\mathrm{}}$

The SI unit of molality is 1 mol/kg

Molarity: It is defined as the number of moles of solute divided by the volume of solution in latice. It is represented by M  The formula of molarity is expressed as

M = $\frac{\mathrm{n}}{\mathrm{V}\mathrm{}\mathrm{}}$

Here, n is the number of moles of solute and V is the volume of solution expressed in liters. The SI unit of molarity is mol/ L .

Mole fraction (x ) : The mole fraction of a substance is defined as the number of moles of that substance divided by the total number of moles of all substances present in the solution.

Mass percent: it is defined as the weight of the solute divided by the total weight of the solution and the obtained result is multiplied by 100 in order to give a percent. Thus, mole fraction and mass percent are unitless.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (A), (D)

Law of conservation of mass: According to this law, the mass is neither created nor destroyed during a chemical reaction in an isolated system. It can only be transformed from one form to another. Burning of wood is an example of conservation of mass. Law of definite proportion:

 According to this law, a compound contains exactly the same elements in the fixed proportion by mass. For example, pure water consists of 11.196 H and 88.9% o by mass.

Law of multiple proportion: According to the law of multiple proportion, when two elements react to form two or more than two chemical compounds, the ratio between different masses of one of the elements combining with a fixed mass of the other is always in the ratio of tiny numbers. For example, compounds of carbon and oxygen. Avogadro's law: This law states that the number of moles of a gas is directly related to the volume enclosed by the gas at constant pressure and temperature.

This is a Matching Type Questions as classified in NCERT Exemplar

(i)- (b); (ii) - (c); (iii) - (a); (iv) - (e); (v) - (d)

(i) 88 g of CO₂

The number of moles is given by the following formula,

Moles = $\frac{\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}}{\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}}$  - (1)

So, the number of moles of CO₂ is calculated by using equation (1) as follows

Moles of CO₂ = $\frac{88\mathrm{}\mathrm{g}}{44\mathrm{}\mathrm{g}\mathrm{}/\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{}\mathrm{}\mathrm{}}$ = 2 mol

Thus, option (i) from column I is matched with (b) from column II

(ii)  1 mol of H₂O gives 6.022*10²³  molecules. So, 6.022*10²³  molecules contain 1 mol of H₂O . Thus, option (ii) from column I is matched with (b) from column II

(iii) 5.6 liters of O₂ at STP

1 mol of gas occupies 22.4 liters of O2 . For 5.6 liters of O₂, the number of moles of oxygen gas is calculated as

Moles of O₂ = $\frac{88\mathrm{}\mathrm{g}}{22.4\mathrm{}\mathrm{L}\mathrm{}\mathrm{}\mathrm{}}$  * 5.6 L = 0.25 mol

(iv) 96 g of O₂

The number of moles of O₂ is calculated by using equation (1) as follows,

Moles of O₂ = $\frac{96\mathrm{}\mathrm{g}}{32\mathrm{}\mathrm{g}\mathrm{}/\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{}\mathrm{}}$  = 3 mol

Thus, option (iv) from column I is matched with (e) from column II.

(v) 1 mol of any gas 1 mole of any gas contains 6.022*10²³  molecules.

Thus, option (v) from column I is matched with (d) from column II.

This is a Matching Type Questions as classified in NCERT Exemplar

(i) → (e) (ii) → (d) (iii) → (b) (iv) → (g) (v) → (c), (h) (vi) → (f) (vii) → (a) (viii) → (i)

Explanation:

(i) It is defined as the number of moles of solute dissolved in 1 litre of solution. So, the unit of molarity is mol L-1
(ii) Mole fraction is defined as the number of moles of a constituent divided by the total number of moles. So, it is unitless.

(iii) Mole is the unit to measure the amount of an atom or molecule in SI system. it is symbolized as "mol".

(iv) It is defined as the number of moles of solute dissolved in 1 kg of solvent, so, the unit of morality is mol kg-1.

(v) It is defined as the force per unit area. so, the unit of pressure is Pascal.

(vi) It is the amount of visible light released per unit solid angle in unit time. so, it's unit will be Candela.

(vii) It is defined as the mass of a substance (in kg) divided by the volume of that substance (in mL). The unit of density is g ml-1

(viii) Mass is used to measure the amount of a substance or an object. The unit of mass is kg.

This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

(i) Both A and R are true and R is the correct explanation of A.

Explanation: The molecular formula gives the actual number of different atoms present in a molecule and the empirical formula gives a simple whole number ratio of different atoms present in a molecule.

Since ethene C₂H₄ can be further simplified, the empirical formula becomes CH₂. The molecular mass of C₂H₄ is 28 g/mol and the empirical mass of CH2 is 14 g/mol.

Hence, the empirical mass of ethene is half of its molecular mass.

This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

Option (B)

Atomic mass unit is defined as a mass that is equal to accurately  the mass of carbon-12 atom.

When scientists compare the relative atomic masses of the elements to the mass of carbon, they find that they are near to a whole number value.

So, both X and B are true but R is the correct explanation of A.

This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

Option (C)

In the decimal portion, only trailing or final zeros are significant. This is in accordance with the rules of significant figures.

The given number, 0.200 has 3 significant figures whereas the number, 200 has only one significant figure because zeros present on the right side or at the end of the decimal number are significant and it is provided that they are present on the right side of the decimal point.

Also, zeros are not significant in numbers without decimals. Thus, X is true but R is false.

This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

Option (C) The combustion reaction of methane is given below

CH₄ + O₂ ->  CO₂ + H₂O

In this reaction, water and carbon dioxide are formed.

According to the reaction, 1 mole of methane gives 2 moles of water. So, 16 g of methane gives 36 g of water

Thus, X is false but R is true

Number of sigma bonds are 10.
[Structure showing 10 sigma bonds in the molecule]

ΔH? = ΔH? + ΔH?
= 2.8 + 98.2 = 101 kJ/mole

C? H? + 13/2 O? → 4CO? + 5H? O
1 mole C? H? (58 g) produces 5 mole H? O (90 g)
∴ 90 g H? O obtained from 58 g C? H?
∴ 72g H? O obtained from (58/90) × 72g = 46.4 g
= 464 × 10? ¹g

P (CO? ) = K? X (CO? )
X (CO? ) = P (CO? )/K? = 0.835 / (1.67 × 10³) = 0.5 × 10? ³
X (CO? ) = n (CO? )/ (n (CO? ) + n (H? O) ≈ n (CO? )/n (H? O) (since n (CO? ) << n (H? O)
n (H? O) in 0.9L = 900g/18gmol? ¹ = 50 mol
n (CO? ) = X (CO? ) × n (H? O) = 0.5 × 10? ³ × 50 = 25 × 10? ³ moles = 25 mmol

Cr? O? ²? + 6Fe²? + 14H? → 2Cr³? + 6Fe³? + 7H? O
n-factor for Cr? O? ²? = 6, for Fe²? = 1
M.E Cr? O? ²? = M.E Fe²?
M? V? n? = M? V? n?
0.02 × 15 × 6 = M? × 10 × 1
M? = (0.02 × 15 × 6)/10 = 0.18 M = 18 × 10? ² M

Fe? ³ + e? → Fe? ² E° = 0.77V
Zn (s) → Zn? ² + 2e? ; E° = 0.76V
Cell reaction: 2Fe? ³ + Zn → 2Fe? ² + Zn? ² E°cell = 1.53V
Ecell = E°cell - (0.059/2)log ( [Zn? ²] [Fe? ²]²/ [Fe? ³]²)
1.5 = 1.53 - (0.06/2)log (1 × [Fe? ²]²/ [Fe? ³]²)
-0.03 = -0.03 log ( [Fe? ²]/ [Fe? ³])²
1 = log ( [Fe? ²]/ [Fe? ³])² => [Fe? ²]/ [Fe? ³] = 10
Let total iron = T. [Fe? ³] + [Fe? ²] = T. [Fe? ³] + 10 [Fe? ³] = T. 11 [Fe? ³] = T.
fraction of Fe? ³ = [Fe? ³]/T = 1/11 ≈ 0.09
This solution seems to differ from the image. Let's follow the image's steps.
log ( [Fe? ²]/ [Fe? ³])² = 1 => ( [Fe? ²]/ [Fe? ³])² = 10
[Fe? ²]/ [Fe? ³] = √10 = 3.16. Then [Fe? ³]/ [Fe? ²] = 1/√10 = 0.316.
fraction of [Fe? ³] = [Fe? ³]/ ( [Fe? ²] + [Fe? ³]) = 1/ (1 + [Fe? ²]/ [Fe? ³]) = 1/ (1+√10) = 1/4.16 = 0.24 = 24 × 10? ²

A + B? 2C
Initial: 1, 1
At eq: 1-x, 1+2x
K = [C]²/ ( [A] [B]) = (1+2x)²/ (1-x)² = 100
(1+2x)/ (1-x) = 10
1+2x = 10-10x => 12x = 9 => x = 3/4
[C] = 1+2x = 1+2 (3/4) = 1+1.5 = 2.5M = 25 × 10? ¹M

Energy per second = 1000 J / 10 s = 100 J/s
Energy of one photon E = hc/λ = (6.626×10? ³? × 3×10? ) / (400×10? ) = 4.965 × 10? ¹? J
Number of electrons ejected = Total energy / Energy per photon = 100 / (4.965 × 10? ¹? ) = 20.14 × 10¹? ≈ 2 × 10²?

PV = nRT
n = PV/RT = (1 atm × 4×10? L) / (0.083 LatmK? ¹mol? ¹ × 300 K) = 1.6 × 10? mol
Mass = n × Molar Mass = 1.6 × 10? mol × 16 g/mol = 25.6 × 10? g ≈ 26 × 10? g

CrCl? ·3NH? ·3H? O gives 3 moles of AgCl precipitate. This means all three Cl? are outside the coordination sphere.
The complex is [Cr (NH? )? (H? O)? ]Cl?
The 3 chloride ions satisfy only primary valency.
secondary valency satisfied by chloride ion = 0