48.67 % (w/W) HCl stock solution has a density of $1.5 g m L^{- 1} .$ The molecular weight of HCl is $36.5 g m o l^{- 1} .$ The volume (mL) of stock required to prepare 200 mL of 0.4 M HCl is: ___

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Solution

Mass of HCl in 1.0 mL stock solution = $1.5 \times \frac{48.67}{100} = 0.73005 g$
Mass of HCl required for 200 mL 0.4 M HCl = $\frac{200}{1000} \times 0.4 \times 36.5 = 0.08 \times 36.5 g$
$∴$ 0.7301 g of HCl is present in 1.0 mL stock solution.
$0.08 \times 36.5 g H C l$ will be present in $\frac{0.08 \times 36.5}{0.7301} \approx 4.00 m L$

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