Chemistry NCERT Exemplar Solutions Class 11th Chapter One

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

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

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