Discussion Forum

29.2% (w/W) HCl, stock solution has a density of 1.25 g mL^-1. The molecular weight of HCl is 36.5 g mol^-1. The volume (mL) of stock solution required to prepare a 200 mL solution 0.4 M HCl is

Answer:

Explanation:

Mass of HCl in 1.0 mL stock solution

   $=1.25 \times \frac{29.2}{100}$=0.365 g

Mass of HCl required for 200 mL  0.4 M HCl

= $\frac{200}{100} \times 0.4 \times 36.5=0.08 \times 36.5g$

$\therefore$   0.365g of HCl is present on 1.0mL stock solution

$\therefore$   $0.08 \times 36.5$ g HCl will be present in $\frac{0.08 \times 36.5}{0.365}=8.0mL$