Calculate the molality of $1 L$ solution of $80 % H_{2} S O_{4} (w / v)$ , Given that density of solution is $1.80 g m l^{- 1}$

8.16

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8.6

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1.02

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10.8

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Solution

The correct option is A $8.16$
The sample of $H_{2} O_{2}$ is $80 %$ by volume
$∴ W_{H_{2} S O_{4}} = 800 g$ in $1000 m L$
Volume of solution $= 1000 m L$
Weight of solution $= 1000.0 \times 1.80$
$= 1800 g$
Weight of water $= 1800 - 800 = 1000 g$
Molality $= \frac{800 \times 1000}{98 \times 1000} = 8.16 m o l k g^{- 1}$

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