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Chemistry NCERT Exemplar Solutions Class 11th Chapter One

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Chemistry NCERT Exemplar Solutions Class 11th Chapter One Class 11th

Which of the following solutions have the same concentration?

(i) 20 g of NaOH in 200 mL of solution

(ii) 5 mol of KCl in 200 mL of solution

(iii) 40 g of NaOH in 100 mL of solution

(iv) 20 g of KOH in 200 mL of solution

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alok kumar singh

Contributor-Level 10

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

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

The molarity (M ) is given by the formula:

M= $\frac{n}{V_{L}}$

On substituting the values in the above equation:

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

= 2.5 mol L^-¹

(B) The molarity is given by the formula:

M= $\frac{n}{V_{L}}$

On substituting the values in the above equation:

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

= 2.5 mol L^-¹

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Chemistry NCERT Exemplar Solutions Class 11th Chapter One Class 11th

Which of the following pairs have the same number of atoms?

(A) 16g of O₂ and 4g of H2

(B) 16g of O₂ and 44g of CO₂

(c) 28g of N₂and 32g of O₂

(D) 12g of c and 23g of Na

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alok kumar singh

Contributor-Level 10

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

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

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

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

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

On substituting the values in the equation (2)

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

number of atoms = 0.5 $*$ 6.022 $*$ 10²³

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

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

= 2 mol

On substituting the values in the equation (2):

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

number of atoms = 2 $*$ 6.022 $*$ 10²³

(B) The number of moles o

...more

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

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

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

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

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

On substituting the values in the equation (2)

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

number of atoms = 0.5 $*$ 6.022 $*$ 10²³

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

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

= 2 mol

On substituting the values in the equation (2):

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

number of atoms = 2 $*$ 6.022 $*$ 10²³

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

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

= 1 mol

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

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

number of atoms = 1 $*$ 6.022 $*$ 10²³

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

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

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

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

number of atoms = 1 $*$ 6.022 $*$ 10²³

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

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

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

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

On substituting the values in the equation (2)

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

number of atoms = 1 $*$ 6.022 $*$ 10²³

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

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

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

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

number of atoms = 1 $*$ 6.022 $*$ 10²³

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

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

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

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

number of atoms = 1 $*$ 6.022 $*$ 10²³

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Chemistry NCERT Exemplar Solutions Class 11th Chapter One Class 11th

Sulphuric acid reacts with sodium hydroxide as follows :

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

When 1L of 0.1M sulphuric acid solution is allowed to react with 1L of 0.1M sodium hydroxide solution, the amount of sodium sulphate formed and its molarity in the solution obtained is

(i) 1 mol L^–1

(ii) 10 g

(iii) 025 mol L^–1

(iv) 55 g

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alok kumar singh

Contributor-Level 10

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 $\frac{0.1}{2}$ = 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 $\frac{0.1}{2}$ mole of Na₂SO₄

No. of mole = $\frac{g i v e n m a s s}{m o l a r m a s s ({N a}_{2} {S O}_{4})}$

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

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

given mass = 7.10 g

Volume of solution after mixing is 2 L.

So, the molarity of Na₂SO₄is

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

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Chemistry NCERT Exemplar Solutions Class 11th Chapter One Class 11th

In the following questions two or more options may be correct.

One mole of oxygen gas at STP is equal to _______.

(i) 6.022 * 10²³molecules of oxygen

(ii) 6.022 * 10²³atoms of oxygen

(iii) 16 g of oxygen

(iv) 32 g of oxygen

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alok kumar singh

Contributor-Level 10

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

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

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

= 0.5 mol

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

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

On substituting the values in the above equation:

0.5 m

...more

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

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

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

= 0.5 mol

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

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

On substituting the values in the above equation:

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

number of molecules = 0.5 $*$ 6.022 $*$ 10²³

The number of moles is given by the following formula,

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

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

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

=1 mol

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

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

On substituting the values in the above equation:

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

Number of molecules = 1 $*$ 6.022 $*$ 10²³

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Chemistry NCERT Exemplar Solutions Class 11th Chapter One Class 11th

Which of the following statements indicates that law of multiple proportion is being followed.

(i) Sample of carbon dioxide taken from any source will always have carbon and oxygen in the ratio 1:2.

(ii) Carbon forms two oxides namely CO₂ and CO, where masses of oxygen which combine with fixed mass of carbon are in the simple ratio 2:1.

(iii) When magnesium burns in oxygen, the amount of magnesium taken for the reaction is equal to the amount of magnesium in magnesium oxide formed.

(iv) At constant temperature and pressure 200 mL of hydrogen will combine with 100 mL oxygen to produce 200 mL of water vapour.

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alok kumar singh

Contributor-Level 10

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

...more

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Chemistry NCERT Exemplar Solutions Class 11th Chapter One Class 11th

Which of the following reactions is not correct according to the law of conservation of mass

(i) 2Mg(s) + O₂(g) → 2MgO(s)

(ii) C₃H₈(g) + O₂(g) → CO₂(g) + H₂O(g)

(iii) P₄(s) + 5O₂(g) → P₄O₁₀(s)

(iv) CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O (g)

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alok kumar singh

Contributor-Level 10

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

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Chemistry NCERT Exemplar Solutions Class 11th Chapter One Class 11th

Which of the following statements is correct about the reaction given below:

4Fe(s) + 3O₂(g) → 2Fe₂O₃(g)

(i) Total mass of iron and oxygen in reactants = total mass of iron and oxygen in product therefore it follows law of conservation of mass.

(ii) Total mass of reactants = total mass of product; therefore, law of multiple proportions is followed.

(iii) Amount of Fe2O3 can be increased by taking any one of the reactants (iron or oxygen) in excess.

(iv) Amount of Fe2O3 produced will decrease if the amount of any one of the reactants (iron or oxygen) is taken in excess.

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alok kumar singh

Contributor-Level 10

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

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Chemistry NCERT Exemplar Solutions Class 11th Chapter One Class 11th

Which of the following statements about a compound is incorrect?

(i) A molecule of a compound has atoms of different elements.

(ii) A compound cannot be separated into its constituent elements by physical methods of separation.

(iii) A compound retains the physical properties of its constituent elements.

(iv) The ratio of atoms of different elements in a compound is fixed.

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

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.

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Chemistry NCERT Exemplar Solutions Class 11th Chapter One Class 11th

If the density of a solution is 3.12 g mL^–1, the mass of 1.5 mL solution in significant figures is _______.

(i) 4.7g

(ii) 4680 * 10^–3g

(iii) 4.680g

(iv) 46.80g

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (A)

The density (d ) is given by the formula:

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

Here, m is the mass and v is the volume

On substituting the values in the above equation:

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

The mass will be 4.7 upto 2 significant figure.

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Chemistry NCERT Exemplar Solutions Class 11th Chapter One Class 11th

The empirical formula and molecular mass of a compound are CH₂O and 180 g respectively. What will be the molecular formula of the compound?

(i) C₉H₁₈O₉

(ii) CH₂O

(iii) C₆H₁₂O₆

(iv) C₂H₄O₂]

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (C)

The empirical formula mass of CH₂O is 30 g

The relation between empirical and molecular formula is given as,

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

On substituting the values in equation (1)

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

Thus, the molecular formula of the compound will be,

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

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