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Concentrated of aqueous sulphuric acid is $98 \% \mathrm{H}_2 \mathrm{SO}_4$ by mass and has a density of $1.80 \mathrm{~g} \mathrm{~mL}^{-1}$ Volume of acid required to make one litre of $0.1 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4$ solution is:
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The correct answer is:
$5.55 \mathrm{~mL}$
Molarity of $\mathrm{H}_2 \mathrm{SO}_4$ solution
$=\frac{98 \times 1000}{98 \times 100} \times 1.84=18.4$
Suppose $\mathrm{V} \mathrm{ml}$ of this $\mathrm{H}_2 \mathrm{SO}_4$ is used to prepare 1 . of $0.1 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4$
$\begin{aligned}
& \therefore \mathrm{V} \times 18.02=1000 \times 0.1 \\
& \mathrm{~V}=\frac{1000 \times 0.1}{18.02}=5.55 \mathrm{ml}
\end{aligned}$
$=\frac{98 \times 1000}{98 \times 100} \times 1.84=18.4$
Suppose $\mathrm{V} \mathrm{ml}$ of this $\mathrm{H}_2 \mathrm{SO}_4$ is used to prepare 1 . of $0.1 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4$
$\begin{aligned}
& \therefore \mathrm{V} \times 18.02=1000 \times 0.1 \\
& \mathrm{~V}=\frac{1000 \times 0.1}{18.02}=5.55 \mathrm{ml}
\end{aligned}$
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