A sample of $H_{2} S O_{4}$ (density 1.8 gram per ml) is 90% by weight. What is the volume of acid that has to be used to make 1 liter of 0.2 M $H_{2} S O_{4}$ ?

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11.2

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12.1

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22.4

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Solution

The correct option is C 12.1
Given that,
Density of acid soln $=$ 1.8 g/ mL.

1000 mL volume will weight 1800g.

Mass of $H_{2} S O_{4}$ present $= 1800 \times \frac{90}{100} = 1620 g$

Number of mole of $H_{2} S O_{4} = \frac{a c t u a l m a s s}{m o l a r m a s s} = \frac{1620}{98} = 16.53 m o l$

Thus molarity of stock solution is 16.53 M.

To prepare 1000 mL of 0.2 M solution,

$M_{1} V_{1} = M_{2} V_{2}$

$16.53 \times x = 0.2 \times 1000$

$x = \frac{200}{16.53}$

$x = 12.099 m l$

$x = 12.1 m l$

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