Question

$29.2\% (w/w) HCl$ 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 of $0.4\, M HCl$ is

Answer 1

Stock solution of $HCl =29.2 \%( w / w )$
Molarity of stock solution of $HCl =\frac{29.2 \times 1000 \times 1.25}{36.5 \times 100}$
If volume of stock solution required $= V \,ml$
$ V \times \frac{29.2}{36.5} \times \frac{1000}{80}=200 \times 0.4 $
$\Rightarrow V =8 \,ml$