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Q.

The volume of concentrated sulphuric acid ( 98 mass % $H_{2} {SO}_{4}$ ,density 1.84 g ${cm}^{- 3})$ required to prepare $5 {dm}^{3}$ of $0.5 mol {dm}^{- 5}$ solution of sulphuric acid is

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a

$204 {cm}^{3}$

b

$68 {cm}^{3}$

c

$136 {cm}^{3}$

d

$272 {cm}^{3}$

answer is B.

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Detailed Solution

$V = \frac{m (100 / 98)}{ρ} = \frac{(n_{2} M_{2}) (100 / 98)}{ρ} = \frac{(M V) M_{2} (100 / 98)}{ρ}$

$= \{(0.5 mol {dm}^{- 3}) (5 {dm}^{3}) (98 g {mol}^{- 1}) (100 / 98)\} / (1.84 g {cm}^{- 3}) = 136 {cm}^{3}$

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