Question

29.2% (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 $200ml$ solution of $0.4$ M $HCl$ is:

• A. $8ml$
• B. $7ml$
• C. $6ml$
• D. $5ml$

Answer 1

The correct option is A $8ml$

$29.2$% (w/w) of $HCl⟹$ $29.2$g of $HCl$ is present in $100$g of solution.

Volume of $100$g of solution $=$ $\frac{100}{1.25}=80$ ml.

$\therefore$ Molarity of given $HCl$ solution $=$ $\frac{\frac{29.2}{36.5}\times 1000}{80}$ $=$ $10$M

$M_1{V}_1=M_2{V}_2$

$10\times V_1$ $=$ $0.4\times 200\times {10}^{-3}$

$V_1=8$ ml