Some Basic Concepts of Chemistry - Result Question 104
####42. $29.2 %(w / W) HCl$ stock solution has density of $1.25 g mL$ ${ }^{-1}$. The molecular weight of $HCl$ is $36.5 g mol^{-1}$. The volume $(mL)$ of stock solution required to prepare a 200 $mL$ solution $0.4 M HCl$ is
(2012)
Subjective Questions
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Solution:
- Mass of $HCl$ in $1.0 mL$ stock solution
$$ =1.25 \times \frac{29.2}{100}=0.365 g $$
Mass of $HCl$ required for $200 mL 0.4 M HCl$
$$ =\frac{200}{1000} \times 0.4 \times 36.5=0.08 \times 36.5 g $$
$\therefore 0.365 g$ of $HCl$ is present in $1.0 mL$ stock solution.
$0.08 \times 36.5 g HCl$ will be present in $\frac{0.08 \times 36.5}{0.365}=8.0 mL$
