Paragraph: Properties such as boiling point, freezing point and vapour pressure of a pure solvent change…
Paragraph:
Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Applications of colligative properties are very useful in day-to-day life. One of its examples is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator automobiles.
A solution $M$ is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixtrue is $0.9$
Given Freezing point depression constant of water $\left(k^{\text {water }}\right)=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$
Freezing point depression constant of ethanol $\left(k_f^{\text {ethanol }}\right)=2.0 \mathrm{Kkg} \mathrm{mol}^{-1}$
Boiling point elevation constant of water $\left(k_b^{\text {water }}\right)=0.52 \mathrm{Kkg} \mathrm{mol}^{-1}$
Boiling point elevation constant of ethanol $\left(k_b^{\text {ethanol }}\right)=1.2 \mathrm{Kkg} \mathrm{mol}^{-1}$
Standard freezing point of water $=273 \mathrm{~K}$
Standard freezing point of ethanol $=155.7 \mathrm{~K}$
Standard boiling point of water $=373 \mathrm{~K}$
Standard boiling point of ethanol $=351.5 \mathrm{~K}$
Vapour pressure of pure water $=328 \mathrm{~mm}$ of $\mathrm{Hg}$
Vapour pressure of pure ethanol $=40 \mathrm{~mm}$ of $\mathrm{Hg}$
Molecular weight of water $=18 \mathrm{~g} \mathrm{~mol}^{-1}$
Molecular weight of ethanol $=46 \mathrm{~g} \mathrm{~mol}^{-1}$
In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative.Question:
The freezing point of the solution $M$ is
$268.7 \mathrm{~K}$
$268.5 \mathrm{~K}$
$234.2 \mathrm{~K}$
$150.9 \mathrm{~K}$
Solution
Solution $M$ is mixture of ethanol and water.
Mole fraction of ethanol is $0.9 \Rightarrow$ Solvent is $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$
Mole fraction of water is $0.1 \Rightarrow \mathrm{H}_2 \mathrm{O}$ is solute
Molality of $\mathrm{H}_2 \mathrm{O}=\frac{n_2}{n_1 M_1}=\frac{0.1}{0.9 \times 46} \times 1000=2.415$
$
\Delta T_f=k_f \cdot m=2 \times 2.415=4.83
$
or Freezing point of solution $=155.7-4.83=150.87 \mathrm{~K}$