JEE MAIN - Chemistry (2024 - 29th January Morning Shift - No. 22)
A solution of $$\mathrm{H}_2 \mathrm{SO}_4$$ is $$31.4 \% \mathrm{H}_2 \mathrm{SO}_4$$ by mass and has a density of $$1.25 \mathrm{~g} / \mathrm{mL}$$. The molarity of the $$\mathrm{H}_2 \mathrm{SO}_4$$ solution is _________ $$\mathrm{M}$$ (nearest integer)
[Given molar mass of $$\mathrm{H}_2 \mathrm{SO}_4=98 \mathrm{~g} \mathrm{~mol}^{-1}$$]
Answer
4
Explanation
$$\begin{aligned}
& M=\frac{n_{\text {solute }}}{V} \times 1000 \\
& =\frac{\left(\frac{31.4}{98}\right)}{\left(\frac{100}{1.25}\right)} \times 1000 \\
& =4.005 \approx 4
\end{aligned}$$
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