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1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constant of benzene is 5.12 K kg mol–1. Find the molar mass of the solute.

1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constant of benzene is 5.12 K kg mol–1. Find the molar mass of the solute.

Grade:11

2 Answers

Gaurav
askIITians Faculty 164 Points
6 years ago
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Rishi Sharma
askIITians Faculty 646 Points
one year ago
Dear Student,
Please find below the solution to your problem.

Freezing point depression constant of benzene = Kf = 5.12 K kg/mol
(Given)
W1 = 50g (Given)
W2 = 100g (Given)
Freezing point of benzene = ΔTf = 0.40K (Given)
Thus,
ΔTf × Kf × W2 × 1000/ W1 × M2
M2 = Kf × W2 × 1000/ W1 × Tf
M2 = 5.12 × 100 × 1000/ 50 × 0.4
M2 = 25600g
Therefore, the molar mass of the solute is 25600g

Thanks and Regards

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