A solution of 2.5 g of ethanoic acid in 75 g of benzene is prepared. What is the molar mass of ethanoic acid, moles of ethanoic acid, mass of benzene and molality of ethanoic acid?
(A) $0.556\, mol\, kg^{-1}$
(B) $0.0417\, mol$
(C) $60\, g\, mol^{-1}$
(D) $0.075\, kg$
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (C), (B), (D), (A)
1. Molar Mass of Ethanoic Acid ($CH_3COOH$)
Ethanoic acid consists of 2 Carbon atoms, 4 Hydrogen atoms, and 2 Oxygen atoms.
- Carbon ($C$): $12.0 \times 2 = 24.0$
- Hydrogen ($H$): $1.0 \times 4 = 4.0$
- Oxygen ($O$): $16.0 \times 2 = 32.0$
- Total: $24 + 4 + 32 = \mathbf{60\text{ g mol}^{-1}}$ (Match: C)
2. Moles of Ethanoic Acid
The number of moles is calculated by dividing the given mass of the solute by its molar mass.
$\text{Moles} = \frac{\text{Given Mass}}{\text{Molar Mass}} = \frac{2.5\text{ g}}{60\text{ g mol}^{-1}} \approx \mathbf{0.0417\text{ mol}}$
(Match: B)
3. Mass of Benzene (Solvent) in kg
Since molality requires the mass of the solvent to be in kilograms, we convert the given 75 g of benzene.
$\text{Mass in kg} = \frac{75\text{ g}}{1000} = \mathbf{0.075\text{ kg}}$
(Match: D)
4. Molality of Ethanoic Acid ($m$)
Molality is defined as the number of moles of solute per kilogram of solvent.
$m = \frac{\text{Moles of Solute}}{\text{Mass of Solvent (kg)}} = \frac{0.0417\text{ mol}}{0.075\text{ kg}} \approx \mathbf{0.556\text{ mol kg}^{-1}}$
(Match: A)