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

Explanation

Molar mass of `C_(2)H_(4)O_(2) : 12 xx 2 +1 xx 4 + 16 xx 2 = 60 " g mol"^(-1)`
Moles of `C_(2)H_(4)O_(2) = (2.5 g)/(60 " g mol"^(-1)) = 0.0417 ` mol
Mass of benzene in kg = 75/ 1000 g `kg^(-1) = 75 xx 10^(-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)`