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Q.

$29.2 % (w / w) HCl$ stock solution has a density of $1.25 {gmL}^{- 1}$ . 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 $0.4 MHCl$ is:

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answer is 8.

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Detailed Solution

Let volume of HCl required $= V_{mL}$

$1.25 \times V \times \frac{29.2}{100} \times \frac{1}{36.5} = \frac{200 \times 0.4}{1000} \Rightarrow V = 8 mL$

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