Question

The freezing point of a solution containing 50 cm of ethylene glycol in 50 g of water is found to be $-{{4}^{0}}C$. Assuming ideal behaviour, the density of ethylene glycol is:
(${{K}_{f}}\,water = 1.86K\,kg\,mo{{l}^{-1}}$)
A. $1.133g\,c{{m}^{-3}}$
B. $4.433g\,c{{m}^{-3}}$
C. $3.235\,g\,c{{m}^{-3}}$
D. none of these

Answer 1

Hint: The concept of depression in freezing point due to addition of non volatile solute is to be used in this question. Substitute the values in the equation and use the relation of mass volume and density to obtain the final answer.

Complete step by step answer:
- In order to answer our question, we need to learn about the depression of freezing point in addition to non volatile solute. Freezing point of a substance is the temperature at which solid and liquid phases of the substance coexist. Actually it is the temperature at which its liquid and solid phases have the same vapour pressure.
- The freezing point of a pure liquid is constant. Now, if we dissolve a non-volatile solute in the pure liquid to constitute a solution, there occurs a lowering in the freezing point. The freezing point of solution refers to the temperature at which the vapour pressure of the solvent in two phases, i.e., liquid solution and solid solvent is the same. Since, the vapour pressure of solvent in solution is lowered, it becomes equal to that of the solid solvent at a lower temperature.
- If ${{K}_{f}}$ is the molal depression constant and $\Delta {{T}_{f}}$ is the depression in freezing point, and m is molality then we have the relation $\Delta {{T}_{f}}={{K}_{f}}\times m$. Now, let us come to the question. We have been given the mass of water as 50 gram and ${{K}_{f}}$ as $1.86K\,mo{{l}^{-1}}\,kg$, $\Delta {{T}_{f}}={{34}^{0}}C$. O, we will put all the values in the equation and obtain:
\[\begin{align}
 & \Delta {{T}_{f}}=\dfrac{1000\times {{K}_{f}}\times w}{m\times W} \\
 & \Rightarrow 34=\dfrac{1000\times 1.86\times 50\times d}{62\times 50};\,as\,\,density=\dfrac{mass}{volume} \\
 & So,d = 1.133g\,c{{m}^{-3}} \\
\end{align}\]
The correct option is option “A” .

Note: It is to be noted that the real formula for finding depression in freezing point is $\Delta {{T}_{f}} = i\times {{K}_{f}}\times m$, where I is Henry’s constant. But for non electrolytes, i = 1 hence it is ignored.