Chemistry - Mole Concept Question with Solution | TestHub

$\underset{(s)}{{\mathrm{CaCO}}_3}+\underset{50\mathrm{mL},0.5\mathrm{M}}{2\mathrm{HCl}(\mathrm{aq})}⟶\underset{(\mathrm{aq})}{{\mathrm{CaCl}}_2}+\underset{(g)}{{\mathrm{CO}}_2}+\underset{(l)}{2{\mathrm{H}}_2\mathrm{O}}$ the number of moles of HCl taken $=0.5\times 0.05$ $=0.025$ moles As we can see that from the above balanced equation is one mole of ${\mathrm{CaCO}}_3(s)$ requires 2 moles of $\mathrm{HCl}(\mathrm{aq})$ therefore, for 0.025 moles of $\mathrm{HCl}(\mathrm{aq}),0.0125$ moles of ${\mathrm{CaCO}}_3(s)$ will be required. Mass of 0.0125 moles of ${\mathrm{CaCO}}_3(s)=0.0125\times$ molar mass of ${\mathrm{CaCO}}_3$ $\begin{array}{rl}& =0.0125\times 100\\ & =1.25\mathrm{g}\end{array}$ But purity of ${\mathrm{CaCO}}_3(s)$ is $95\%$. Therefore, the actual amount of ${\mathrm{CaCO}}_3(s)$ required is Percentage purity $=\frac{\text{ weight of pure substance }}{\text{ weight of impure sample }}\times 100$ $95=\frac{1.25}{\text{ weight of impure sample }}\times 100$ weight of impure sample $=\frac{1.25\times 100}{95}$ $=1.32\mathrm{g}$ Hence. option 1 is correct.