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98%(w/w) $H_2SO_4$ stock solution has a density of 1.8g.m$L^{-1}$.The molecular weight of $H_2SO_4$ is 98g$mol^{-1}$.The volume (ml) of stock solution required to prepare a 450ml of 1.0M $H_2SO_4$ is

$\begin{array}{1 1}(a)\;10ml\\(b)\;20ml\\(c)\;25ml\\(d)\;30ml\end{array}$

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Answer: 25 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 $H_2SO_4 = \large\frac{W}{M} $$ \times\large\frac {1000}{V\;mL}$ $ = \large\frac{98 \times 1000 \times 1.8 }{98 \times 100}$$ = 18\;M$
The volume of stock solution required to prepare a $450\;mL$ of $1\;M\; H_2SO_4$ is $V_1=\large\frac{M_2\;V_2}{V_1}$
$\Rightarrow V_1 = \large\frac{1 \times 450}{18}$$ = 25\;mL$
answered Feb 13, 2014 by sreemathi.v
edited Jul 16, 2014 by balaji.thirumalai

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