Page No 25: - Chapter 1 Some Basic Concepts Of Chemistry class 11 ncert solutions Chemistry - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Question 1.32:

Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes:

Molar mass of argonAnswer:

= 39.947 gmol^–1

Question 1.33:

Calculate the number of atoms in each of the following (i) 52 moles of Ar (ii) 52 u of He (iii) 52 g of He.

Answer:

(i) 1 mole of Ar = 6.022 × 10²³ atoms of Ar

52 mol of Ar = 52 × 6.022 × 10²³ atoms of Ar

= 3.131 × 10²⁵ atoms of Ar

(ii) 1 atom of He = 4 u of He

Or,

4 u of He = 1 atom of He

1 u of He  atom of He

52u of He   atom of He

= 13 atoms of He

(iii) 4 g of He = 6.022 × 10²³ atoms of He

52 g of He  atoms of He

= 7.8286 × 10²⁴ atoms of He

Question 1.34:

A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate (i) empirical formula, (ii) molar mass of the gas, and (iii) molecular formula.

Answer:

(i) 1 mole (44 g) of CO₂ contains 12 g of carbon.

3.38 g of CO₂ will contain carbon

= 0.9217 g

18 g of water contains 2 g of hydrogen.

 0.690 g of water will contain hydrogen 

= 0.0767 g

Since carbon and hydrogen are the only constituents of the compound, the total mass of the compound is:

= 0.9217 g + 0.0767 g

= 0.9984 g

Percent of C in the compound 

= 92.32%

Percent of H in the compound 

= 7.68%

Moles of carbon in the compound 

= 7.69

Moles of hydrogen in the compound = 

= 7.68

Ratio of carbon to hydrogen in the compound = 7.69: 7.68

= 1: 1

Hence, the empirical formula of the gas is CH.

(ii) Given,

Weight of 10.0L of the gas (at S.T.P) = 11.6 g

Weight of 22.4 L of gas at STP 

= 25.984 g

≈ 26 g

Hence, the molar mass of the gas is 26 g.

(iii) Empirical formula mass of CH = 12 + 1 = 13 g

n = 2

Molecular formula of gas = (CH)_n

= C₂H₂

Question 1.35:

Calcium carbonate reacts with aqueous HCl to give CaCl₂ and CO₂ according to the reaction, CaCO₃₍s₎ + 2 HCl₍aq₎ → CaCl₂₍aq₎ + CO₂₍g) + H₂O₍l₎

What mass of CaCO₃ is required to react completely with 25 mL of 0.75 M HCl?

Answer:

0.75 M of HCl ≡ 0.75 mol of HCl are present in 1 L of water

≡ [(0.75 mol) × (36.5 g mol^–1)] HCl is present in 1 L of water

≡ 27.375 g of HCl is present in 1 L of water

Thus, 1000 mL of solution contains 27.375 g of HCl.

Amount of HCl present in 25 mL of solution

= 0.6844 g

From the given chemical equation,

2 mol of HCl (2 × 36.5 = 73 g) react with 1 mol of CaCO₃ (100 g).

Amount of CaCO₃ that will react with 0.6844 g =

10073×0.6844= 0.9375 g

Question 1.36:

Chlorine is prepared in the laboratory by treating manganese dioxide (MnO₂) with aqueous hydrochloric acid according to the reaction

4HCl₍aq₎ + MnO₂₍s₎ → 2H₂O₍l₎ + MnCl₂₍aq₎ + Cl₂₍g₎

How many grams of HCl react with 5.0 g of manganese dioxide?

Answer:

1 mol [55 + 2 × 16 = 87 g] MnO₂ reacts completely with 4 mol [4 × 36.5 = 146 g] of HCl.

5.0 g of MnO₂ will react with

of HCl

= 8.4 g of HCl

Hence, 8.4 g of HCl will react completely with 5.0 g of manganese dioxide.