Question

2% (w/w) HCl stock solution has a density of $1.25gm{L}^{-1}$. The molecular weight of HCl is $36.5gmo{l}^{-1}$. The volume (mL) of stock solution required to prepare 200 mL solution of 0.4 M HCl is _____.

Answer 1

2% (w/w) HCl solution means 2 gm of HCl present in 100 gm of solution.

Moles of HCl $=\frac{2}{36.5}$

Density = 1.25 $gm{L}^{-1}$

$\therefore$ Volume of solution $=\frac{Mass}{Density}$

$=\frac{100}{1.25}$

$=80mL$

Molarity of stock solution of HCl,

$=\frac{Moles\times 1000}{V\left(mL\right)}$

$=\frac{2\times 1000}{36.5\times 80}=0.684M$

The volume of stock solution required to prepare $200mL$ of $0.4M$ HCl is,

$V_1=\frac{M_2{V}_2}{V_1}$

$=\frac{0.4\times 200}{0.684}=116.9mL$

Hence, the correct answer is 116.9 $mL$.