The molality of solution obtained by dissolving 1.5 g of ethanoic acid $(CH_3COOH)$ in 25g of benzene is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $1.0\, mol\, kg^{-1}$
Molality (m) = moles of solute / mass of solvent in kg
Moles of Ethanoic Acid
Molar mass of CH₃COOH = C₂H₄O₂ = 2 ×12+4×1+2×16=60g/mol
Moles = 1.5 / 60 =0.025 mol
Mass of Solvent in kg
25 g =0.025 kg
Molality
m =0.025 / 0.025 =1.0 mol/kg = $1.0\, mol\, kg^{-1}$