JEE MAIN - Chemistry (2023 - 25th January Morning Shift - No. 14)
The density of a monobasic strong acid (Molar mass 24.2 g/mol) is 1.21 kg/L. The volume of its solution required for the complete neutralization of 25 mL of 0.24 M NaOH is __________ $$\times$$ 10$$^{-2}$$ mL (Nearest integer)
Answer
12
Explanation
$$
\begin{aligned}
& \text { millimole of } \mathrm{NaOH}=0.24 \times 25 \\\\
& \therefore \text { millimole of acid }=0.24 \times 25 \\\\
& \Rightarrow \text { mass of acid }=0.24 \times 25 \times 24.2 \mathrm{mg} \\\\
& \text { for pure acid, } \\\\
& \mathrm{V}=\frac{\mathrm{w}}{\mathrm{d}} ;(\mathrm{d}=1.21 \mathrm{~kg} / \mathrm{L}=1.21 \mathrm{~g} / \mathrm{ml}) \\\\
& \therefore \mathrm{V}=\frac{0.24 \times 25 \times 24.2}{1.12} \times 10^{-3} \\\\
& =120 \times 10^{-3} ~\mathrm{ml} \\\\
& =12 \times 10^{-2} ~\mathrm{ml}
\end{aligned}
$$
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