Question

$30gm$ of ethanoic acid present in $100gm$ of water, determine molality of ethanoic acid in water?

Answer 1

The molar mass of ethanoic acid $C{H}_3-COOH=60$ g/mol.
The number of moles of ethanoic acid $=\frac{\text{30 g}}{\text{60 g/mol}}=\text{0.5 mol}$
Mass of water $=\text{100 g}=\frac{\text{100 g}}{\text{1000 g/kg}}=\text{0.100 kg}$
The molality of ethanoic acid $=\frac{\text{0.5 mol}}{\text{0.100 kg}}=\text{5 mol/kg}$