Question

Mole fraction of a nonelectrolyte in aqueous solution is 0.07 if kf is 1.860 mole per kg depression in freezing point delta t f is

Answer 1

Answer:- depression in freezing point of the solution is 7.77 degree C.

Solution:- From given info, mole fraction of the non electrolyte in aqueous solution is 0.07.

Sum of the mole fractions is always 1. So, the mole fraction of the solvent(water) = 1 - 0.07 = 0.93

From mole fractions, there are 0.07 moles of solute and 0.93 moles of solvent.

kf = 1.860 mole per kg

$\Delta Tf=?$

The formula used for this type of problems is:

$\Delta Tf=i*m*kf$

where m is the molality of the solution and i is Van't hoff factor. Since the solute is non electrolyte, the value of i is 1 as it does not ionize to give the ions.

molality is moles of solute per kg of solvent.

mass of solvent(water) = 0.93*18.0 = 16.74 g = 0.01674 kg

molality = $\frac{0.07mol}{0.01674kg}$

molality = 4.18m

Let's plug in the values in the formula to calculate depression in freezing point:

$\Delta Tf=1*4.18*1.860$

$\Delta Tf$ = 7.77

So, depression in freezing point of the solution is 7.77 degree C or kelvin.