Question

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

Answer 1

Moles of `C_(2)H_(4)O_(2):12xx2+1xx4+16xx2=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 is 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")=(0.0417xx" mol "xx1000" g kg"^(-1))/(75"g")`
`=0.556" mol kg"^(-1)`