Question

Correct order of freezing point of given solution 0.1M glucose 0.2M Urea 0.1M NaCl 0.05MCaCl2

Answer 1

Kf is the molal freezing point depression constant of the solvent (1.86 °C/m for water). m = molality = moles of solute per kilogram of solvent.

Answer 2

Answer : The correct order is, $0.2\text{M }urea=0.1\text{M }NaCl>0.05\text{M }CaCl_2>0.1\text{M }glucose$

Explanation :

Formula used for lowering in freezing point :

$\Delta T_f=i\times k_f\times m$

where,

$\DeltaT_f$ = change in freezing point

$k_f$ = freezing point constant

m = molality

i = Van't Hoff factor

As we know that the boiling point depends on the molality and the Van't Hoff factor.

Now we have to calculate the Van't Hoff factor for the given solutions.

(a) $C_6H_{12}O_6$, glucose is a non-electrolyte solute that means they retain their molecularity, and not undergo association or dissociation.

So, Van't Hoff factor = 1

The freezing point will be :

$\Delta T_f=i\times k_f\times m=1\times k_f\times 0.1=0.1k_f$

(b) Urea is a non-electrolyte solute that means they retain their molecularity, and not undergo association or dissociation.

So, Van't Hoff factor = 1

The freezing point will be :

$\Delta T_f=i\times k_f\times m=1\times k_f\times 0.2=0.2k_f$

(C) The dissociation of 0.1 M $NaCl$ will be,

$NaCl\rightarrow Na^++Cl^-$

So, Van't Hoff factor = Number of solute particles = $Na^++Cl^-$ = 1 + 1 = 2

The freezing point will be :

$\Delta T_f=i\times k_f\times m=2\times k_f\times 0.1=0.2k_f$

(D) The dissociation of 0.05 M $CaCl_2$ will be,

$CaCl_2\rightarrow Ca^{2+}+2Cl^-$

So, Van't Hoff factor = Number of solute particles = $Ca^{2+}+2Cl^-$ = 1 + 2 = 3

The freezing point will be :

$\Delta T_f=i\times k_f\times m=3\times k_f\times 0.05=0.15k_f$

From this we conclude that, the order of freezing point will be:

$0.2\text{M }urea=0.1\text{M }NaCl>0.05\text{M }CaCl_2>0.1\text{M }glucose$