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29.2%(w/w) HCl stock solution has a density of 1.25g$ml^{-1}$.The molecular weight of HCl is 36.5$gmol^{-1}$.The volume (ml) of stock solution required to prepare a 500ml of 0.1M HCl is

$\begin{array}{1 1}(a)\;5ml\\(b)\;15ml\\(c)\;20ml\\(d)\;25ml\end{array}$

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Answer: 5 mL
The number of moles of solute in the concentrated solution is equal to the number of moles in the dilute solution $\rightarrow M_1\;V_1 = M_2\;V_2$ where 1 and 2 stand for the concentrated and diluted solutions.
$\Rightarrow$ Molarity of stock solution of $HCl = \large\frac{W}{M} $$ \times\large\frac {1000}{V\;mL}$ $ = \large\frac{29.2\times1000\times1.25}{36.5 \times 100}$$ = 10\;M$
The volume of stock solution required to prepare a $500\;mL$ of $0.1\;M\;HCl$ is $V_1=\large\frac{M_2\;V_2}{V_1}$
$\Rightarrow V_1 = \large\frac{0.1 \times 500}{10}$$ = 5\;mL$
answered Feb 13, 2014 by sreemathi.v
edited Jul 16, 2014 by balaji.thirumalai

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