contestada

A certain substance condenses at a temperature of 123.3°C . But if a 65.0 gm sample of X is prepared with 24.6g of urea ((NH2)2CO) dissolved in it, the sample is found to have a condensation point of 124.3°C instead.
1. Calculate the molal boiling point elevation constant Kb of X . Round your answer to significant digits.

Respuesta :

dsdrajlin

Answer:

Kb = 0.159 °C.kg/mol

Explanation:

The condensation point has the same value than the boiling point. The elevation in the boiling point can be calculated using the following expression.

ΔT = Kb × b

where,

ΔT: elevation in the boiling point

Kb: molal boiling point elevation constant

b: molality of the solute

The molality of the urea is:

[tex]b=\frac{24.6g}{(60.06g/mol).65.0 \times 10^{-3} kg} =6.30mol/kg[/tex]

Then,

ΔT = Kb × b

(124.3°C-123.3°C) = Kb × (6.30 mol/kg)

Kb = 0.159 °C.kg/mol

Otras preguntas