The boiling point of a pure liquid is 353.23 K. If we add 2.70 g of a non-volati

Question

The boiling point of a pure liquid is 353.23 K. If we add 2.70 g of a non-volatile solute in 90 g of liquid, the boiling point of the solution rises to 354.11 K. What will be the molar mass of the non-volatile solute? Take the value of Kb of liquid to be 2.53 K kg mol-1. The boiling point of a pure liquid is 353.23 K. If we add 2.70 g of a non-volatile solute in 90 g of liquid, the boiling point of the solution rises to 354.11 K. What will be the molar mass of the non-volatile solute? Take the value of Kb of liquid to be 2.53 K kg mol-1.

Explanation / Answer

Elevation in boiling point = 1000 * Kb * w / (W*m)

354.11 - 353.23 = 1000 * 2.53 * 2.70 / (90 * m)

m = molar mass non volatile solute = 1000 * 2.53 * 2.70 / (90 * 0.88) = 86.25 g/mol

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