Question

$29.2\%\left(w/w\right)HCl$ stock solution has a density of $1.25gm{L}^{-1}$. The molecular weight of $HCl$ is $36.5gmo{l}^{-1}.$ The volume $\left(mL\right)$ of stock solution required to prepare a $200mL$ solution of $0.4MHCl$ is:

Answer 1

$29.2\%\left(w/w\right)HCl\text{has density}=1.25g/mL$
Now, mole of HCl required in 0.4 HCl
$=0.4\times 0.2$ mole = 0.08 mole
if v mL of original HCl solution is taken then mass of solutions = 1.25 v
mass of $HCl=(1.25v\times 0.292)$
mole of $HCl=\frac{1.25v\times 0.292}{36.5}=0.08$
so, $v=\frac{36.5\times 0.08}{0.292\times 1.25}mL=8mL$