Given, $\left[d_{\mathrm{H}_2 \mathrm{SO}_4}=1.84 \mathrm{~g} / \mathrm{cc}\right]$
$$
\begin{aligned}
& \text { Molality }(m)=\frac{\text { Number of moles of solute }}{\text { Weight of solvent in kg }} . \\
& \begin{aligned}
& \text { Weight }=93 \mathrm{~g} \\
& \text { Molecular weight }=98 \mathrm{~g} \\
& \text { Solute }=\frac{\text { weight }}{\text { mol }} \times \frac{1}{\text { weight of solvent }} \\
& d=\frac{M}{V} m=d \times V \\
& m=1.84 \mathrm{gmL}^{-1} \times 1000 \mathrm{~mL}=1840 \mathrm{~g} \\
& \text { Weight of } \mathrm{H}_2 \mathrm{SO}_4=\frac{93}{100} \times 1000=930 \mathrm{~g}
\end{aligned}
\end{aligned}
$$


$$
\text { So, weight of solvent } \begin{aligned}
& =1840-930 \mathrm{~g} \\
& =910 \mathrm{~g}=0.91 \mathrm{~kg} \\
& =\frac{93 \mathrm{~g}}{98 \mathrm{~g}} \times \frac{1}{0.91 \mathrm{~kg}} \\
& =1.042 \mathrm{~mol} / \mathrm{kg}
\end{aligned}
$$