30 grams of a compound that does not ionize in H_2O are added to 500 grams of H_

Question

30 grams of a compound that does not ionize in H_2O are added to 500 grams of H_2O. It is found that the freezing point of the solution is lowered by 0.450 K relative to the freezing point of pure H_2O. If the solution is ideal and sufficiently dilute to use the limiting form of the freezing point depression equation, the data show the molar mass of the compound to be greater than 300 g mol^-1. between 250 to 300 g mol^-1. between 200 to 250 g mol^-1. between 150 to 200 g mol^-1. less than 150 g mol^-1.

Explanation / Answer

mass of solute = 30 g

mass of solvent = 500g

depression in freezing point = 0.45

Kf of water = 1.86 K Kg/m

We know delat Tf = Kf x m

0.45 = 1.86 x (30/M)(1000/500)

or molar mass M = 248g/mol

Thus the ioption C is correct.

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