Question

Calculate molality of 2.5 grams of Ethanoic Acid $\left(C{H}_{3}COOH\right)$ in 75 grams of Benzene.

Answer 1

The molar mass of ethanoic acid is $2\left(12\right)+2\left(16\right)+4\left(1\right)=60$ g/mol. The mass of ethanoic acid is divided with molar mass of ethanoic acid to obtain number of moles of ethanoic acid.

Number of moles of ethanoic acid $=\frac{2.5g}{60g/mol}=0.04167mol$

The mass of benzene is 75 grams or $\frac{75}{1000}=0.075$ kg.
The molality of ethanoic acid in benzene is the ratio of the number of moles of ethanoic acid to the mass of benzene (in kg).

It is $\frac{0.04167mol}{0.075kg}=0.556m$