Solutions and Colligative Properties 2 Question 34
34. Addition of $0.643 \mathrm{~g}$ of a compound to $50 \mathrm{~mL}$ of benzene (density : $0.879 \mathrm{~g} / \mathrm{mL}$ ) lowers the freezing point from $5.51^{\circ} \mathrm{C}$ to $5.03^{\circ} \mathrm{C}$. If $K_{f}$ for benzene is 5.12 , calculate the molecular weight of the compound.
$(1992,2 \mathrm{M})$
Passage Based Questions
Passage 1
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 of automobiles.
A solution $M$ is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9 .
Given, freezing point depression constant of water
$$ \left(K_{f}^{\text {water }}\right)=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} $$
Freezing point depression constant of ethanol
$$ \left(K_{f}^{\text {ethanol }}\right)=2.0 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} $$
Boiling point elevation constant of water
$$ \left(K_{b}^{\text {water }}\right)=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} $$
Boiling point elevation constant of ethanol
$$ \left(K_{b}^{\text {ethanol }}\right)=1.2 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} $$
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 $=32.8 \mathrm{~mm} \mathrm{Hg}$
Vapour pressure of pure ethanol $=40 \mathrm{~mm} \mathrm{Hg}$
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.
$(2008,3 \times 4 \mathrm{M}=12 \mathrm{M})$
Show Answer
Solution:
- $-\Delta T_{f}=5.51-5.03=0.48$
$\Rightarrow \quad-\Delta T_{f}=0.48=K_{f} \cdot m$
$\Rightarrow \quad 0.48=5.12 \times \frac{0.643}{M} \times \frac{1000}{50 \times 0.879}$
$\Rightarrow \quad M=156 \mathrm{~g} / \mathrm{mol}$
