Calculate molality of 2.5 g of ethanoic acid $[C H_{3} C O O H]$ in 75 g of benzene.

0.551 m o l k g^{- 1}

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0.556 m o l k g^{- 1}

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0.558 m o l k g^{- 1}

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0.552 m o l k g^{- 1}

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Solution

The correct option is B $0.556 m o l k g^{- 1}$

MW of $C H_{3} C O O H = 60 g m o l^{- 1}$

$m = \frac{w_{2} \times 1000}{M w_{2} \times w_{1}} = \frac{2.5 g \times 1000}{60 g m o l^{- 1} \times 75 g} = 0.556 m o l k g^{- 1}$

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