Question

At 293 K, the density of an aqueous solution containing 120 g of urea (NH\(_2\)CONH\(_2\)) per dm\(^3\) is 1.02 kg dm\(^{-3}\). The mole fraction of urea is approximately: 

• A. 0.038
• B. 0.962
• C. 0.02
• D. 0.98

Hint:

Mole fraction calculations require accurate determination of the total number of moles present in the solution, ensuring that contributions from all components are correctly accounted for.

Answer 1

The Correct Option is A

The mole fraction of urea can be calculated as follows: 1. Calculate the mass of the solution: \[ {Density} = \frac{{Mass of solution}}{{Volume of solution}} \] Given that the density is 1.02 kg/dm³ and the volume is 1 dm³: \[ {Mass of solution} = 1.02 \times 1000 = 1020 { g} \] 2. Calculate the moles of urea: The molar mass of urea (NH\(_2\)CONH\(_2\)) is: \[ 14 + (2 \times 1) + 12 + 16 + 14 + (2 \times 1) = 60 { g/mol} \] Thus, the moles of urea: \[ \frac{120}{60} = 2 { moles} \] 3. Calculate the moles of water: The mass of water in the solution: \[ {Mass of water} = 1020 - 120 = 900 { g} \] The molar mass of water is 18 g/mol, so the moles of water: \[ \frac{900}{18} = 50 { moles} \] 4. Calculate the mole fraction of urea: \[ X_{{urea}} = \frac{{moles of urea}}{{moles of urea} + {moles of water}} \] \[ X_{{urea}} = \frac{2}{2 + 50} = \frac{2}{52} = 0.038 \] Thus, the mole fraction of urea in the solution is 0.038.