Question:

Pure benzene freezes at $5.3^{\circ} C$. A solution of $0.223 \,g$ of phenylacetic acid $(C_6H_5CH_2COOH)$ in $4.4\, g$ of benzene $(h_f = 5.12\,kg\,mol^{-1})$ freezes at $4.47^{\circ} C$. From this observation, one can conclude that

Updated On: Jul 28, 2022
  • phenylacetic acid exists as such in benzene
  • phenylacetic acid undergoes partial ionisation in benzene
  • phenylacetic acid undergoes complete ionisation in benzene
  • phenylacetic acid dimerises in benzene
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The Correct Option is D

Solution and Explanation

Depression in freezing point, $\Delta T_{f}=k_{f} m$ (where $m=$ molality) $=5.12 \times \frac{0.223 \times 1000}{4.4 \times 136}$ $=5.12 \times 0.372$ $=1.91^{\circ} C$ $\therefore$ Theoretical $\Delta T_{f}=1.91$ Actual $\Delta T_{f}=5.3^{\circ} C -4.47^{\circ} C$ $=0.83^{\circ} C$ $\therefore i=\frac{0.83}{1.91}=0.43$ $\because i < 1 $ $ \therefore $ Association will take place. Thus, it can be said that phenylacetic acid dimerises in benzene.
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Concepts Used:

Colligative Properties

Colligative Property of any substance is entirely dependent on the ratio of the number of solute particles to the total number of solvent particles but does not depend on the nature of particles. There are four colligative properties: vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure.

Examples of Colligative Properties

We can notice the colligative properties of arrangements by going through the accompanying examples:

  • On the off chance that we add a spot of salt to a glass full of water, its freezing temperature is brought down impressively than its normal temperature. On the other hand, the boiling temperature is likewise increased and the arrangement will have a lower vapor pressure. There are also changes observed in its osmotic pressure.
  • In the same way, if we add alcohol to water, the solution’s freezing point goes down below the normal temperature that is usually observed for either pure alcohol or water.

Types of Colligative Properties

  1. Freezing point depression: ΔTf =1000 x kf x m2 /(M2 x m1)
  2. Boiling point elevation: ΔTb = kb m
  3. Osmotic pressure: π = (n2/V) RT
  4. Relative lowering of vapor pressure: (Po - Ps)/Po