Question

Determine the freezing point depression of a solution that contains 30.7 g glycerin (C3H8O3, molar mass = 92.09 g/mol) in 376 mL of water. Some possibly useful constants for water are Kf = 1.86°C/m and Kb = 0.512°C/m.

Answer 1

Answer : The freezing depression of a solution is $-1.65^oC$

Explanation :  Given,

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

Mass of glycerin (solute) = 30.7 g

Volume of water (solvent) = 376 mL

Molar mass of glycerin = 92.09 g/mole

First we have to calculate the mass of water.

As we know that, the density of water is 1.0 g/mL.

So, mass of water = Density of water × Volume of water

Mass of water = 1.0 g/mL × 376 mL

Mass of water = 376 g = 0.376 kg   (1 g = 0.001 kg)

Now we have to calculate the freezing point depression of a solution.

Formula used :  

$\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{\text{Mass of glycerin}}{\text{Molar mass of glycerin}\times \text{Mass of water in Kg}}$

where,

$\Delta T_f$ = change in freezing point

$\Delta T_s$ = freezing point of solution = ?

$\Delta T^o$ = freezing point of water = $0^oC$

i = Van't Hoff factor = 1  (for glycerin non-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

$0^oC-T_s=1\times (1.86^oC/m)\times \frac{30.7g}{92.09g/mol\times 0.376kg}$

$T_s=-1.65^oC$

Therefore, the freezing depression of a solution is $-1.65^oC$