Question

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

Answer 1

Molality (m)`=("Mass of " CH_(3)COOH//"Molar mass of " CH_(3)COOH)/("Mass of bezene in kg")`
Mass of `CH_(3)COOH =2.5g`.
Molar mass of `CH_(3)COOH=2xx12+4xx1+2xx16=60" g mol"^(-1)`.
`"Mass of solvent(benzene)"=75g=(75g)/(1000)=0.075kg`.
Molality(M)`=((2.5)//(60.0mol^(-1)))/((0.075kg))=0.556" mol kg"^(-1)=0.556m`