Question

Calculate boiling point depression and freezing point depression. Make sure to include correctly written dissociation equation.

A. 0.25 m ethylene glycol -C2H6O2

B. 0.25 m NaCl

Answer 1

Answer A

Boiling point Elevation

delta Tb =i Kb *m

delta Tb is elevation in boiling point

Kb is boiling point elevationconstant

m is molality of solution which is given as 0.25 m

We have to calculate i for ethylene glycol

Ethylene glycol is an organic compound so it does not dissociate in solution and has an i factor of 1.

So i would be 1

delta Tb=Kb *m

Kb is constant(= Kb = 0.512°C/m)

delta Tb=i 0.25 * 0.512

=1* 0.25 *0.512

=0.128 deg CELCIUS

delta Tf (Depression in freezing point)= KfFreezing point depression constant) * m(molality)

Kf= 1.86°C/m

delta Tf=i 0.25 * 1.86 deg C

=1 *0.25 * 1.86

=0.465 deg Celcius

Answer B)

For 0.25 molal NaCl

delta Tf(Elevation in boiling point for NaCl)=Kf(molal boiling point elevation constant)

delta Tf = i Kf*0.25

For an electrolyte, such as sodium chloride, you must take into consideration that if 1 mol of NaCl dissolves, 2 mol of particles would result (1 mol Na⁺, 1 mol Cl^–).

NaCl....>Na+ + Cl-

Therefore, the van’t Hoff factor should be 2

i is vant Hoff factor

delta Tf=2 * Kf *0.25

Kf= the molal freezing point constant (1.86 °C/m)

delta Tf=2 *1.86 *0.25

=0.93 deg Celcius

delta Tb=i * Kb *m

Kb is molal boiling point elevation constant

(= Kb = 0.512°C/m)

=2 * 0.512 * 0.25

=0.256 degC