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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Solutions
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$29.2\%$(w/w) $HCl$ stock solution has a density of $1.25gmL^{-1}$. The molecular weight of $HCl$ is $36.5g\;mol^{-1}.$ Find the volume $(ml)$ of stock solution required to prepare a $200\;ml$ of $0.4\;M\; HCl$ is

$\begin{array}{1 1}(a)\;4ml\\(b)\;6ml\\(c)\;8ml\\(d)\;10ml\end{array}$

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Answer: 8 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 $200\;mL$ of $0.4\;M\;HCl$ is $V_1=\large\frac{M_2\;V_2}{V_1}$
$\Rightarrow V_1 = \large\frac{0.4 \times 200}{10}$$ = 8\;mL$
answered Feb 13, 2014 by sreemathi.v
edited Jul 16, 2014 by balaji.thirumalai
 

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