Some Basic Concepts of Chemistry 1 Question 21
42. $29.2 %(w / W) \mathrm{HCl}$ stock solution has density of $1.25 \mathrm{~g} \mathrm{~mL}$ ${ }^{-1}$. 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 $0.4 \mathrm{M} \mathrm{HCl}$ is
(2012)
Subjective Questions
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Solution:
- Mass of $\mathrm{HCl}$ in $1.0 \mathrm{~mL}$ stock solution
$$ =1.25 \times \frac{29.2}{100}=0.365 \mathrm{~g} $$
Mass of $\mathrm{HCl}$ required for $200 \mathrm{~mL} 0.4 \mathrm{M} \mathrm{HCl}$
$$ =\frac{200}{1000} \times 0.4 \times 36.5=0.08 \times 36.5 \mathrm{~g} $$
$\therefore 0.365 \mathrm{~g}$ of $\mathrm{HCl}$ is present in $1.0 \mathrm{~mL}$ stock solution.
$0.08 \times 36.5 \mathrm{~g} \mathrm{HCl}$ will be present in $\frac{0.08 \times 36.5}{0.365}=8.0 \mathrm{~mL}$
