OnlineCBT
Chemistry · PYQ NEET UG 2022
हिन्दीWhat mass of \( 95\% \) pure \( \text{CaCO}_3 \) will be required to neutralise \( 50\text{ mL} \) of \( 0.5\text{ M HCl} \) solution according to the following reaction? \( \text{CaCO}_{3(s)} + 2\text{HCl}_{(aq)} \rightarrow \text{CaCl}_{2(aq)} + \text{CO}_{2(g)} + 2\text{H}_2\text{O}_{(l)} \) [Calculate upto second place of decimal point]
Correct answer · Option 4
\( 1.32\text{ g} \)
Explanation
Moles of \( \text{HCl} = \frac{50 \times 0.5}{1000} = 0.025\text{ mol} \). Moles of \( \text{CaCO}_3 \) required = \( 0.0125\text{ mol} \). Mass = \( 0.0125 \times 100 = 1.25\text{ g} \). Pure mass = \( \frac{1.25}{0.95} = 1.32\text{ g} \).
