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

Explanation

Molar mass of `C_(2)H_(2)O_(2) : 12xx2+1xx4+16xx2`
`= 60 g mol^(-1)`
Moles of `C_(2)H_(4)O_(2)=(2.5g)/(60 g mol^(-1))=0.0417 mol`
Mass of benzene in `kg = 75 g//1000 g kg^(-1)`
`= 75xx10^(-3)kg`
Molality of `C_(2)H_(4)O_(2)=("Moles of " C_(2)H_(4)O_(2))/("kg of benzene")`
`= (0.0417 mol xx 1000 g kg^(-1))/(75 g)`
`= 0.556 mol kg^(-1)`