(solved)Question 1.35 of NCERT Class XI Chemistry Chapter 1

$\displaystyle \underline{\underline{\text{Step 1: Calculate mols of HCl}}}$

by def. $\displaystyle \frac{\text{no. of mols (n)}}{\text{vol. in liters}} = \text{Molarity}$

Given vol. of HCl = 25 mL = 0.025 L

$\displaystyle \therefore \frac{\text{n}}{0.025 \text{ L}} = 0.75 \text{ M}$

$\displaystyle \implies n = 0.75 \times 0.025$

$\displaystyle \therefore n = 0.0188 \text{ mol . . . (1)}$

$\displaystyle \underline{\underline{\text{Step 2: Calculate mols of CaCO}_3}}$

the balanced equation is:

$\displaystyle \text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{CO}_2 + \text{H}_2\text{O}$

2 mol HCl completely reacts with 1 mol CaCO3

by (1) we have 0.0188 mol HCl

$\displaystyle \therefore$ 0.0188 mol reacts with $\displaystyle \frac{1}{2} \times 0.0188$

$\displaystyle = 0.0094 \text{ mol CaCO}_3$

$\displaystyle \underline{\underline{\text{Step 3: Calculate mass of CaCO}_3}}$

Molar mass of CaCO3 = 40 + 12 + (3 x 16) = 100 g/mol

we know, $\displaystyle \frac{w}{M} = n$

put M = 100 g/mol, n = 0.0094 mol

$\displaystyle \therefore w = 0.0094 \times 100 = 0.94\text{ g}\:\underline{Ans}$