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 (integer)
12
Solution
$$
\begin{aligned}
& \text { millimole of } \mathrm{NaOH}=0.24 \times 25 \\\\
& \therefore \text { millimole of acid }=0.24 \times 25 \\\\
& \Rightarrow \text { mass of acid }=0.24 \times 25 \times 24.2 \mathrm{mg} \\\\
& \text { for pure acid, } \\\\
& \mathrm{V}=\frac{\mathrm{w}}{\mathrm{d}} ;(\mathrm{d}=1.21 \mathrm{~kg} / \mathrm{L}=1.21 \mathrm{~g} / \mathrm{ml}) \\\\
& \therefore \mathrm{V}=\frac{0.24 \times 25 \times 24.2}{1.12} \times 10^{-3} \\\\
& =120 \times 10^{-3} ~\mathrm{ml} \\\\
& =12 \times 10^{-2} ~\mathrm{ml}
\end{aligned}
$$
About this question
Subject: Chemistry · Chapter: Redox Reactions · Topic: Oxidation States
This question is part of PrepWiser's free JEE Main question bank. 49 more solved questions on Redox Reactions are available — start with the harder ones if your accuracy is >70%.