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$29.2 \%(\mathrm{w} / \mathrm{w}) \mathrm{HCl}$ stock solution has a density of $1.25 \mathrm{gmL}^{-1}$. The molecular weight of $\mathrm{HCl}$ is $36.5 \mathrm{~g} \mathrm{~mol}^{-1}$. The volume $(\mathrm{mL})$ of stock solution required to prepare a 200 $\mathrm{mL}$ solution of $0.4 \mathrm{M} \mathrm{HCl}$ is :
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Verified Answer
The correct answer is:
8
Molarity of stock solution of $\mathrm{HCl}$
$=\frac{29.2 \times 1000 \times 1.25}{100 \times 36.5}$
Let the volume of stock solution required $=V \mathrm{~mL}$
Thus, $V \times \frac{29.2 \times 1000 \times 1.25}{100 \times 36.5}=200 \times 0.4$ $\Rightarrow V=8 \mathrm{~mL}$
$=\frac{29.2 \times 1000 \times 1.25}{100 \times 36.5}$
Let the volume of stock solution required $=V \mathrm{~mL}$
Thus, $V \times \frac{29.2 \times 1000 \times 1.25}{100 \times 36.5}=200 \times 0.4$ $\Rightarrow V=8 \mathrm{~mL}$
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