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 of 0.4 M HCl is

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answer is 8.

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Detailed Solution

$\begin{array}{l} M = \frac{(\frac{W}{W}) % \times d \times 10}{M_{solute}} = \frac{29 .2 \times 1 .25 \times 10}{36 .5} = 10 M \\ M_{1} V_{1} = M_{2} V_{2}; V_{1} = \frac{M_{2} V_{2}}{M_{1}} = \frac{200 \times 0 .4}{10} = 8 ml \end{array}$

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