Calculate the molality of 25 g of ethanoic acid ($CH_3COOH$) in 250 g benzene

$1.664\, mol\, kg^{-1}$

The correct answer is Option (2) → $1.664\, mol\, kg^{-1}$

Formula for Molality (m): $m=\frac{\text{moles of solute}}{\text{mass of solvent (in kg)}}$

Calculate Moles of Solute (Ethanoic Acid, $\text{CH}_3\text{COOH)}$:

$\text{Molar mass of CH}_3\text{COOH:}$

$\text{C}=12.01\,\text{g/mol}$

$\text{H}=1.008\,\text{g/mol}$

$\text{O}=16.00\,\text{g/mol}$

$\text{Molar mass}=2(12.01)+4(1.008)+2(16.00)=24.02+4.032+32.00=60.052\,\text{g/mol}$

$\textbf{Moles}=\frac{\text{mass}}{\text{molar mass}}$

$\text{Moles of CH}_3\text{COOH}=\frac{25\,\text{g}}{60.052\,\text{g/mol}}\approx0.4163\,\text{mol}$

Convert Mass of Solvent to kg:

$\text{Mass of benzene}=250\,\text{g}=0.250\,\text{kg}$

$\textbf{Calculate Molality:}m=\frac{0.4163\,\text{mol}}{0.250\,\text{kg}}=1.6652\,\text{mol kg}^{-1}$ = $1.664\, mol\, kg^{-1}$