Question

HCl stock solution has a density of $1.25$ g $m{L}^{-1}$. The molecular volume (mL) of stock solution required to prepare a $200$ mL solution.

Answer 1

The number of moles of solute in the concentrated solution is equal to the number of moles in the dilute solutions.

Since,

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

where $1$ and $2$ stand for the concentrated and dilute solutions.

Implies that

Molarity of stock solution of $HCl=\frac{W}{M}\times \frac{1000}{VmL}$

$=\frac{29.2\times 1000\times 1.25}{36.5\times 100}$

The volume of stock solution required to prepare a $200mL$of $0.4MHCl$ is $V_1=\frac{M_2{V}_2}{V_1}$

Implies that

$V_1=\frac{0.4\times 200}{10}=8mL$