Question

[ncchem 3.2.052A.] Some physical properties of water are shown below. freezing point 0.00°C boiling point 100.00°C Kf 1.86°C/m Kb 0.512°C/m What is the freezing point of a solution of glucose, a nonelectrolyte, that contains 66.0 g of C6H12O6 dissolved in 373 g of water? Use molar masses with at least as many significant figures as the data given. °C What is the boiling point of the solution? °C

Answer 1

Solution :

We have to use following equation to find freezing point of the solution.

Delta Tf = m x Kf

Here delta Tf is the depression in freezing point.

m is molality , kf is the freezing point constant .

The solute given in this problem is non electrolyte.

Calculation of m (molality) of the solute

Molality = moles of solute / volume of solute in Kg

Here water is solvent and we need to get its mass in kg

Conversion of mass into kg

Mass of water in kg = 373 g x 1 kg / 1000 g = 0.373 kg

Now calculation of moles of glucose

Moles of glucose = 66.0 g x 1 mol / 180.1559 g = 0.36634 mol

Molality of glucose = 0.36634 mol / 0.373 kg = 0.9823 m

Now we find delta Tf (freezing point depression)

Delta Tf = 0.9823(mol/kg) x (1.86 ⁰C/ m )=1.82

Delta Tb is 1.82. This shows that the freezing point is lowered by 1.82 deg C

So the freezing point of solution

= 0 deg C (pure solvent) – Delta T =-1.82 deg C

Calculation of boiling point of the solution:

We use the following formula

Delta Tb = m x kb

Here Delta Tb is the elevation in boiling point .

Here kb is boiling point constant

Lets use these values and get the delta Tb

Delta Tb = 0.9823 (mol / kg ) x (0.512 deg C /m)

=0.50287 deg C

We found delta Tb.

This is elevated T.

T of solution = Boiling T (pure solvent ) + Delta Tb = 100 .00 ⁰ C +0.50287 ⁰C

= 100.50287 ⁰ C