Some Basic Concepts of Chemistry

This is a Short Answer Type Questions as classified in NCERT Exemplar

(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.

This is a Short Answer Type Questions as classified in NCERT Exemplar

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.

This is a Short Answer Type Questions as classified in NCERT Exemplar

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.

This is a Long Answer Type Questions as classified in NCERT Exemplar

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.

This is a Long Answer Type Questions as classified in NCERT Exemplar

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