A sulphuric acid solution contains $98 %$ $(m / v) H_{2} S O_{4}$ . The density of the solutions is $1.98 g c m^{3}$ . The molarity and molatity respectively of the solution are:

19.8 M

500 m

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10 M

1.5 m

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10 M

10 m

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1.0 M

1 m

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Solution

The correct option is A $19.8 M$ & $500 m$
Given:

Density=1.98g
weight of $H_{2} S O_{4}$ =98g
weight of solution=100g

$∴$ volume of solution= $\frac{100}{1.98} = 50.5 c m^{3}$

weight of solvent(water)=100-98=2g
Molecular weight of $H_{2} S O_{4}$ =98g/mol

$∴$ no of moles in 98g= $\frac{g i v e n m a s s}{m o l a r m a s s} = \frac{m}{w} = \frac{98}{98} = 1 m o l$

Molality = $\frac{n o . o f m o l e s o f s o l u t e}{w e i g h t o f s o l v e n t i n g r a m} X 1000$

$= \frac{1}{2} X 1000 = 500 m o l / k g$

Molarity = $\frac{n o o f m o l e s o f s o l u t e}{v o l u m e o f s o l u t i o n i n m l} X 1000$

$= \frac{1}{50.5} X 1000 = 19.8 m o l / L H_{2} S O_{4}$ is the answer.

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