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Commercially available $\mathrm{H}_2 \mathrm{SO}_4$ is $98 \mathrm{~g}$ by weight of $\mathrm{H}_2 \mathrm{SO}_4$ and $2 \mathrm{~g}$ by weight of water. It's density is $1.38 \mathrm{~g} \mathrm{~cm}^{-3}$. Calculate the molality $(m)$ of $\mathrm{H}_2 \mathrm{SO}_4$ (molar mass of $\mathrm{H}_2 \mathrm{SO}_4$ is $98 \mathrm{~g} \mathrm{~mol}^{-1}$ )
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Verified Answer
The correct answer is:
$500 \mathrm{~m}$
Given weight of solute $98 \mathrm{~g}$, molar mass of solute $98 \mathrm{~g} \mathrm{~mol}^{-1}$ weight of solute $=2 \mathrm{~g}$
$$
\begin{aligned}
\text { Molality } & =\frac{\text { weight of } \mathrm{H}_2 \mathrm{SO}_4}{\text { molecular weight of } \mathrm{H}_2 \mathrm{SO}_4} \\
& \times \frac{1000}{\text { weight of water }} \\
& =\frac{98}{98} \times \frac{1000}{2}=500 \mathrm{~m}
\end{aligned}
$$
$$
\begin{aligned}
\text { Molality } & =\frac{\text { weight of } \mathrm{H}_2 \mathrm{SO}_4}{\text { molecular weight of } \mathrm{H}_2 \mathrm{SO}_4} \\
& \times \frac{1000}{\text { weight of water }} \\
& =\frac{98}{98} \times \frac{1000}{2}=500 \mathrm{~m}
\end{aligned}
$$
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