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98%(w/w) $H_2SO_4$ stock solution has a density of 1.8gm$L^{-1}$.The molecular weight of $H_2SO_4$ is 98g$mol^{-1}$.The volume(ml) of stock solution required to prepare a 1800ml of 1.0N $H_2SO_4$ is

$\begin{array}{1 1}(a)\;5ml\\(b)\;50ml\\(c)\;25ml\\(d)\;250ml\end{array}$

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Answer: 50 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$
Normality of stock solution of $H_2SO_4 = 2 \times 18M = 36 N$
$M_1V_1 = M_2 V_2$ can be expressed as $\rightarrow N_2V_2 = N_1V_1$
$\Rightarrow V_2 = \large\frac{N_1V_1}{N_2}$$ = \large\frac{1 \times 1800}{36}$$ = 50\; mL$
answered Feb 13, 2014 by sreemathi.v
edited Jul 16, 2014 by balaji.thirumalai
 

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