Question

A solution of 1 moles of sucrose in 1000 grams of water freezes at $-{{0.2}^{\circ }}C$ . How many grams of ice would have separated at $-{{0.4}^{\circ }}C$-
[A] 500 grams
[B] 300 grams
[C] 250 grams
[D] 400 grams

Answer 1

To solve this question use the definition of molality and the molal freezing constant. Establish two relations of molality from each of the mentioned definitions for the first solution and also when it turns to ice. Compare the two to find out the required mass of ice.

Complete step by step solution:
Here, we can use the relation of depression in freezing point and the molality.
The cryoscopic constant gives us a relation between the molality and the depression in freezing point that we can use here. The relation is -
$\Delta {{T}_{f}}={{K}_{f}}\times m$
Here, $\Delta {{T}_{f}}$ is the depression in freezing point and ‘m’ is the molality.
In the given question we have 1 mole of sucrose in 1000 grams of water and it freezes at $-{{0.2}^{\circ }}C$.
We know that molality is the number of moles of solute per kilogram of solvent. Here we have 1 mole of solute for 1000 grams of water. Therefore, we can write that-$m=\dfrac{1}{1}$ (as 1000g = 1 kg).
Also, according to the cryoscopic relation we can write that-$m=\dfrac{0.2}{{{K}_{f}}}$
Now, we have to find how many grams of ice would be separated at $-{{0.4}^{\circ }}C$.
We can write here that molarity = $\dfrac{1}{W}$ .
And according to the cryoscopic relation we have, $m =$ $\dfrac{0.4}{{{K}_{f}}}$
Therefore, combining the two relations of molality from the above equation we can write that-
$m=\dfrac{0.2}{{{K}_{f}}}=1\text{ and m = }\dfrac{0.4}{{{K}_{f}}}=\dfrac{1}{W}$
Now, from these two equations we can write that - $0.4W=0.2$
We will get that W = 0.5Kg = 500g.
Initially we have 1000 g of water.
Therefore, the amount of ice separated as ice = 1000 – 500 = 500g.

Therefore, the correct answer is option [A] 500 grams.

Note: The molal freezing constant is also known as the cryoscopic constant. As cryoscopic constant give us a relation between the molality and the depression in freezing point, similarly there is another constant named ebullioscopic constant which gives us a relation between the molality and the elevation in boiling point and it is denoted as ${{K}_{b}}$ . Through the process of ebullioscopy and cryoscopy, we can calculate the value of unknown molar mass through a known constant.