What mass of 95% pure $CaCO_3$ will be required to neutralise 50 mL of 0.5 M HCl solution according to the following reaction? $CaCO_{3(s)}+2HCl_{(aq)}\rightarrow CaCl_{2(aq)}+CO_{2(g)}+2H_2O_{(l)}$ [Calculate upto second place of decimal point]

$CaCO_3 + 2HCl \rightarrow CaCl_2 + CO_2 + H_2O$ Moles of HCl $= 0.5 \times \frac{50}{1000} = 0.025$ mol Moles of pure $CaCO_3 = \frac{1}{2} \times 0.025 = 0.0125$ mol Mass of pure $CaCO_3 = 0.0125 \times 100 = 1.25$ g % purity $= \frac{wt.\ of\ pure\ substance}{wt.\ of\ impure\ sample} \times 100 \Rightarrow 95 = \frac{1.25}{w} \times 100$ $w = \frac{1.25 \times 100}{95} = 1.32$ g