The molality of a solution made by dissolving 2.5 g of ethanoic acid $(CH_3COOH)$ in 75 g of benzene is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $0.556\, mol\, kg^{-1}$
Molality m is defined as:
$m = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}}$
Step 1: Moles of ethanoic acid
Molar mass of $\mathrm{CH_3COOH} = 60 \,\text{g mol}^{-1}$
$\text{Moles} = \frac{2.5}{60} = 0.0417 \,\text{mol}$
Step 2: Mass of benzene in kg
$75\,\text{g} = 0.075\,\text{kg}$
Step 3: Molality
$m = \frac{0.0417}{0.075} = 0.556 \,\text{mol kg}^{-1}$