Question

Calculate the amount of ice that will separate of cooling a solution containing $$50g$$ of ethylene glycol containing $$200g$$ water to $$-{9.3}^{o}C$$:
A. $$19.35g$$
B.$$28.45g$$
C. $$38.71g$$
D.$$56.9g$$

Answer 1

Hint: Here we have to calculate the depression of freezing point which means, we need to calculate the amount of ice that will separate from cooling. And it is calculated by using the terms, given weight, molecular weight and a constant, $$K_f$$. Where, $$K_f$$ is the Molal freezing point depression constant which indicates the change of freezing of the solvent.

Complete answer:
The amount of ice that will separate from cooling the solution is not equal to$$19.35g$$. Hence, the option (A) is incorrect.
The amount of ice that will separate from cooling the solution is not equal to$$28.45g$$. Hence, the option (B) is incorrect.
Freezing point depression is a colligative property and it is the lowering of the freezing point of the solvent by adding the solute. And it can be calculated by using the equation,
$$\mathrm{\Delta }T={\displaystyle \frac{1000\times K_f\times w}{W\times m}}$$
Where, $$K_f$$ is the molal freezing point depression constant $$=1.86Kmo{l}^{-1}$$
$$w=$$Given weight of ethylene glycol$$=50g$$
W$$=$$Molecular weight of ethylene glycol
$$m=$$Mole of solute$$=1$$
$$\mathrm{\Delta }\mathrm{T}=$$ The freezing point depression is equal to $$-{9.3}^{o}C$$.
Therefore,
$$9.3={\displaystyle \frac{1000\times 1.86\times 50}{62\times W}}$$
By rearranging the equation we will get the value of ‘W’.
$$W=161.29$$
Therefore, the amount of ice separated can be calculated by subtracting the weight of ethylene glycol with the weight of water.
$$iceseparated=200-161.29=38.71g$$
Hence, the amount of ice that will separate from cooling the solution is equal to $$38.71g$$. Thus, the option (C) is correct.
The amount of ice that will separate from cooling the solution is not equal to $$56.9g$$
Hence, the option (D) is incorrect.

Note:
We have to remember that the amount of ice that will separate from cooling the solution is equal to $$38.71g$$. It is calculated by using the equation of freezing point depression. Freezing point depression is a colligative property and it is the lowering of freezing point of the solvent by adding the solute. And it mainly depends upon the constant, $$K_f$$, which is a Molal freezing point depression constant which indicates the change of freezing of the solvent.