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Normality

What is final...

Question

What is final molarity of 30% (w/w) $H_{2} O_{2}$ aqueous solution, having density 1.11 g/ml, if the solution is diluted to triple the initial volume?

9.8 M

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4.85 M

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12.6 M

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3.26 M

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Solution

The correct option is D 3.26 M
30% (w/w) means 30 g of $H_{2} O_{2}$ is present in 100 g of the solution.
Mass of solvent = 100 g - 30 g = 70 g
Number of moles of $H_{2} O_{2} = \frac{30 g}{m o l a r m a s s o f H_{2} O_{2}} = \frac{30 g}{34 g / m o l} = \frac{30}{34} m o l$
Volume of initial solution $(V_{1})$ = $\frac{m a s s o f s o l u t i o n}{d e n s i t y o f s o l u t i o n} = \frac{100 g}{1.11 g / m l} ≃ 90 m l = 0.09 L$
$M o l a r i t y o f i n i t i a l s o l u t i o n (M_{1}) = \frac{n u m b e r o f m o l e s o f H_{2} O_{2}}{v o l u m e o f s o l u t i o n i n L}$
$= \frac{30}{0.09 \times 34} = 9.8 M$

Now, since the number of moles of $H_{2} O_{2}$ remain same, before and after the dilution.
$∴ M_{1} V_{1} = M_{2} V_{2}$ , where $M_{2}$ and $V_{2}$ are final molarity and volume, after dilution, respectively.
$\to M_{2} = \frac{9.8 \times 90}{3 \times 90} ≃ 3.26 M$

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