Question

Commercially available concentrated hydrochloric acid contains $38$ $HCl$ by mass.
$\left(i\right)$ What is the molarity of the solution (density of solution = $1.19gm{L}^{-1}$)?
$\left(ii\right)$ What volume of concentrated $HCl$ is required to make $1.0L$ of an $0.10MHCl?$

Answer 1

$38$ $gm$ $HCl$ is present in $100$ $gm$ solution.

Volume of solution $=\frac{100}{1.19}=84$ $mL$

Moles of $HCl=\frac{38}{36.5}=1.04$ $moles$

$\left(i\right)$ $Molarity=\frac{moles}{Volumeofsolution\left(litres\right)}=\frac{}{1.04\times 1000}84=12.38$ $M$

$\left(ii\right)$ No. of moles of $HCl$ require $=0.1$ $moles$

Mass of $HCl=0.1\times 36.5=3.65$ $gm$

Mass of solution having $3.65$ $gm$ $HCl=y$

$\frac{38}{100}=\frac{3.65}{y}$

$⟹y=9.6$ $gm$

Volume of concentration $HCl$ required $=\frac{9.6}{1.19}=8.067$ $mL$