Question

What mass of $95\%$ pure $CaC{O}_3$ will be required to neutralise $50mL$ of $0.5MHCl$ solution according to the following reaction?
$CaC{O}_{3(s)}+2HC{l}_{(aq)}\to CaC{l}_{2(aq)}+C{O}_{2(g)}+2H_2{O}_{(l)}$ [Calculate upto second place of decimal point]

• A. $9.50g$
• B. $1.25g$
• C. $1.32g$
• D. $3.65g$

Answer 1

Answer: (C)

$CaC{O}_{3(s)}+2HC{l}_{\text{(aq.) }}\to CaC{l}_{2\text{ (aq.) }}+C{O}_{2(a)}+H_2{O}_{(0)}$
no. of moles of $CaC{O}_3$ (pure) $=\frac{1}{2}\times$ mole of $HCl$
$\text{ [Mole }=\text{ molarity }\times \text{ volume(in ltr. })]$
$=\frac{1}{2}\times 0.5\times \frac{50}{1000}=0.0125$
weight of $CaC{O}_3$ (pure) $=$ mole $\times$ mol. wt
$=0.0125\times 100=1.25g$
$\%$ purity $=\frac{\text{ wt. of pure substance }}{\text{ wt. of impure sample }}\times 100$
$95=\frac{1.25}{\text{ wt. of impure sample }}\times 100$
wt. of impure sample $=\frac{1.25\times 100}{95}=1.32g$