Question

A solution is prepared by dissolving 4.66 g of KCl in enough distilled water to give 250 mL of solution. KCl is a strong electrolyte. How will the freezing point of the solution be different from that of pure water?

Answer 1

Answer : The solution will be $0.931^oC$  lower than water.

Explanation :  Given,

Molal-freezing-point-depression constant $(K_f)$ for water = $1.86^oC/m$

Mass of KCl (solute) = 4.66 g

Volume of water = 250 mL

Density of water = 1.00 g/mL

So,

Mass of water (solvent) = $Density\times volume=1.00g/mL\times 250mL=250g=0.250kg$

Molar mass of KCl = 74.5 g/mole

Formula used :  

$\Delta T_f=i\times K_f\times m\\\\\Delta T=i\times K_f\times\frac{\text{Mass of KCl}}{\text{Molar mass of KCl}\times \text{Mass of water in Kg}}$

where,

$\Delta T$ = change in freezing point

i = Van't Hoff factor = 2  (for KCl electrolyte)

$K_f$ = freezing point constant for water = $1.86^oC/m$

m = molality

Now put all the given values in this formula, we get

$\Delta T=2\times (1.86^oC/m)\times \frac{4.66g}{74.5g/mol\times 0.250kg}$

$\Delta T=0.931^oC$

$T^o-T_s=0.931^oC$

From this we conclude that, the solution will be $0.931^oC$  lower than water.

Therefore, the solution will be $0.931^oC$  lower than water.

Answer 2

Answer:

Depression in freezing point = 2 X 1.853 X 0.25 = 0.9625

Thus this will be the difference between the freezing point of pure water and the solution.

Explanation:

On adding any non volatile solute to a solvent its boiling point increases and its freezing point decreases [these are two of the four colligative properties].

The depression in freezing point is related to molality of solution as:

ΔTf$=iK_{f}Xmolality$

where

ΔTf= depression in freezing point

Kf= cryoscopic constant of water = 1.853 K. kg/mol.

i = Van't Hoff factor = 2 ( for KCl)

molality = $\frac{molesofsolute}{massofsolvent(Kg)}$

moles of solute = mass / molarmass = 4.66 / 74.55 =0.0625

mass of solvent = mass of solution (almost)

considering the density of solution to be 1g/mL

mass of solvent = 250 grams = 0.250 Kg

molality = $\frac{0.0625}{0.25}= 0.25$

Putting values

depression in freezing point = 2 X 1.853 X 0.25 = 0.9625

Thus this will be the difference between the freezing point of pure water and the solution.