Question

Calculate the freezing point of a solution containing 60 g of glucose (Molar mass= 180 g/mol) in 250 g of water.

Answer 1

Hint: The freezing point is the temperature at which the solvent in liquid state and solvent in solid state are at equilibrium, so that their vapor pressures are equal.

Complete step by step answer:
In the question it is given that
Weight of glucose $$w_2$$ = 60 g
Weight of water $$w_1$$= 205 g
Molar mass of glucose $$m_2$$ = 180 g/ml
Freezing point of pure water $$K_f$$ = 1.86 k kg/mol
The formula to calculate depression in freezing point of solution is
$$\begin{array}{rl}& \mathrm{\Delta }{T}_f={\displaystyle \frac{K_f\times w_2\times 1000}{m_2\times w_1}}\\ & \text{ = }{\displaystyle \frac{1.86\times 60\times 1000}{180\times 250}}\\ & \text{ = }2.48K\end{array}$$
Formula to calculate freezing point of solution from depression in freezing point is as follows
$$\mathrm{\Delta }{T}_f=T_f^0-T_f$$
We know that $$T_f^0$$= 273.15 K
$$\begin{array}{rl}& \mathrm{\Delta }{T}_f=T_f^0-T_f\\ & T_f=T_f^0-\mathrm{\Delta }{T}_f\\ & \text{ = 273}\text{.15}-2.48\\ & \text{ = 270}\text{.67 K}\end{array}$$

Therefore the freezing point of the solution containing 60 g of glucose (Molar mass= 180 g/mol) in 250 g of water is 270.67 K.

Note: Don’t be confused with the words depression in freezing point and freezing point.
Freezing point: The freezing point is the temperature at which the solvent in liquid state and solvent in solid state are at equilibrium, so that their vapor pressures are equal.
Depression in freezing point: The depression in freezing point is the difference in the freezing points of the solution from its pure solvent. This is true that if any solute is added to a solvent, the freezing point of the solution will be lesser than the freezing point of the pure solvent.