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:

(C), (B), (D), (A)

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)