Question

The freezing point of aqueous solution that contains 5% by mass urea, 1% by mass KCl and 10% by mass of glucose is (\[{{K}_{f\left( {{H}_{2}}O \right)}}=1.86\operatorname{K}{{\operatorname{molality}}^{-1}}\]):
(A) 290.2 K
(B) 285.5 K
(C) 269.48 K
(D) 250 K

Answer 1

Hint: First of all, find the total number of moles of solute present in the solution by using the following formula-
\[No.\text{ of moles}=\dfrac{Given\operatorname{mass}}{Molar\operatorname{mass}}\]
Then use the formula given below, to find the change in the freezing point-
\[\Delta {{T}_{f}}={{K}_{f}}m\]
Subtract the change in temperature to the original freezing point of water.

Complete step by step solution:
We all know that when a pure solution is mixed with solid impurities or impurities that have a higher melting or boiling point than the solvent, the freezing point of the overall solution decreases. This all happens because the vapour pressure of the pure solvent decreases due to the presence of solid solute particles which makes the solution freeze at a lower temperature than it did previously.
The depression in freezing point can be calculated by the following formula-
\[\Delta {{T}_{f}}={{K}_{f}}m\]
Where $\Delta {{T}_{f}}$ is the change in freezing point of the overall solution; ${{K}_{f}}$ is the proportionality constant which is also known as Molal Depression Constant and m is the molality of the solution in question.
It is to be noted that the value of the constant is specific to a particular condition and the solution taken into account during the experiment. Here, as you can see, we already have that value. All we have to do is find the molality of the solution; to do that we have to find the number of moles of each of the constituents present in the solution.
As given in the question, the solution contains 5 g of urea, 1 g of KCl and 10 g of glucose present in 100 g of the solution, as all of these values are in percentage. The molar mass of urea, potassium chloride and glucose is 60 g, 74.55 g and 180 g respectively. So we calculate the number of moles by applying the formula mentioned below:
\[No.\text{ of moles}=\dfrac{Given\operatorname{mass}}{Molar\operatorname{mass}}\]
We get 0.833, 0.055 and 0.0134 as the number of moles of urea, glucose and potassium chloride. Adding all of them, the total number of moles present in the solution is 0.1517. Molality, as you might know, is the number of moles of solute present in 1 Kg of solvent. As the total mass of all the solute is 16 g, the mass of the solvent is-
\[100\operatorname{g}-16\operatorname{g}=84\operatorname{g}\]
That means, 16 g of solute is present in 84 g of solvent or we can also say that 0.1517 moles of solute are present in 84 g of solvent. So,
\[\operatorname{In}1000\operatorname{g}\to \dfrac{0.1517\times 1000}{84}=1.805\operatorname{moles}\text{ of solute}\]
Which by definition means that the solution is 1.805 molal. Now applying the formula for calculating the depression in freezing point of the solution that we mentioned first, we get the following-
\[\begin{align}
& \Delta {{T}_{f}}={{K}_{f}}m=1.86\times 1.805 \\
& \Rightarrow \Delta {{T}_{f}}=3.3573 \\
\end{align}\]
The freezing point of the pure solvent, which in this case is water, is 273.15 K. The change in temperature or $\Delta {{T}_{f}}$ is-
\[\Delta {{T}_{f}}=T{}^\circ -T=273.15-T\]
Above, $T{}^\circ $ is the freezing point of the pure solvent and T is the freezing point of the solution. Now,
\[\begin{align}
& 273.15-T=3.3573 \\
& \Rightarrow T=269.79\operatorname{K} \\
\end{align}\]

So, the solution will freeze at 269.79 K. Out of all the options given to us, the closest is option (C) 269.48 K, which is our answer.

Note: If in the question the solvent is not mentioned, then a student needs to assume that water is the solvent. You should always be aware of the concentration given to us in the question. In this case, it was percent by mass but it could be any other thing such as volume or mass-by-volume. The first task in this type of question is to find a way to calculate the number of moles of the solute