The molarity of a sulphuric acid solution is $2.32 \mathrm{~mol} \mathrm{dm}^{-3}$. If the density of solution is $1.14 \mathrm{~g} \mathrm{~cm}^{-3}$, the molality of the solution will be

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Aicharya · Accepted Answer

For $1 \mathrm{~L}$ of solution, we have $n_{2}=2.32 \mathrm{~mol}$ and $V=1000 \mathrm{~cm}^{3}$
Mass of solution, $m=V ho=\left(1000 \mathrm{~cm}^{3}ight)\left(1.14 \mathrm{~g} \mathrm{~cm}^{-3}ight)=1140 \mathrm{~g}$
Mass of sulphuric acid, $m_{2}=n_{2} M_{2}=(2.32 \mathrm{~mol})\left(98 \mathrm{~g} \mathrm{~mol}^{-1}ight)=227.36 \mathrm{~g}$
Mass of solvent, $m_{1}=m-m_{2}=1140 \mathrm{~g}-227.36 \mathrm{~g}=912.64 \mathrm{~g}=0.913 \mathrm{~kg}$
Molality of solution $=\frac{n_{2}}{m_{1}}=\frac{2.32 \mathrm{~mol}}{0.913 \mathrm{~kg}}=2.54 \mathrm{~mol} \mathrm{~kg}^{-1}$ ^

The molarity of a sulphuric acid solution is $2.32 \mathrm{~mol} \mathrm{dm}^{-3}$. If the density of…

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