Calculate molality of 2.5g of ethanoic a...

Calculate molality of `2.5g` of ethanoic acid `(CH_(3)COOH)` in `75 g` of benzene.

`0.278 "mol kg"^(-1)`

`0.556 " mol kg"^(-1)`

`5.56 "mol kg"^(-1)`

`27.8 "mol kg"^(-1)`

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The correct Answer is:

`m=(W_(B))/(M_(B)).(1000)/(W_(A))`

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