When 5 g of a nonelectrolyte is added to 25 g of water, the new freezing point is -2.5 degrees Celsius. What is the molecular mass of the . . .unknown compound?

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justaguide | College Teacher | (Level 2) Distinguished Educator

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When a substance is dissolved in a solvent it reduces the freezing point of the solvent. The decrease in the temperature at which the liquid freezes is given by a relation derived by Ebbing as dT = Kf*cm, where Kf is a constant for every liquid and cm is the molality of the solution after the impurity has been dissolved.

For water the freezing-point-depression constant is 1.858.

Let the molecular mass of the unknown substance be m, when 5 g of this is added to 25 g of water the molality is (5/m)/25. The new freezing point is -2.5 degrees Celsius, or there is a drop of 2.5 degree Celsius.

This gives dT = 2.5 = 1.858*mc

=> mc = 2.5/1.858 = 1.345

(5/m)/25 = 1.345

=> m = 0.1486 g

This gives the molecular mass of the unknown compound as 0.1486 g

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