The molality of a solution made by dissolving 2.5 g of ethanoic acid $(CH_3COOH)$ in 75 g of benzene is:

$0.556\, mol\, kg^{-1}$

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

$m = \frac{0.0417}{0.075} = 0.556 \,\text{mol kg}^{-1}$