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First Law of Thermodynamics

The ratio of ...

Question

The ratio of the boiling point (in K) of ethanol to its freezing point (in K) is 9:4, while the ratio of its latent heat of vaporization to its latent heat of fusion is 9:1. If the boiling point of pure ethanol is 360 K and freezing point of a solution of a compound P in ethanol is 152 K, then what is boiling point of same solution?

355.5 K

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364.5 K

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368 K

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352 K

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Solution

The correct option is B 364.5 K
Given, $\frac{T_{b}}{T_{f}} = \frac{9}{4} a n d \frac{l_{v}}{l_{f}} = \frac{9}{1}$
So, $\frac{k_{b}}{k_{f}} = \frac{\frac{R T_{b}^{2}}{1000 l_{v}}}{\frac{R T_{f}^{2}}{1000 l_{f}}} = \frac{T_{b}^{2}}{T_{f}^{2}} \times \frac{l_{f}}{l_{b}} = \frac{9^{2}}{4^{2}} \times \frac{1}{9} = \frac{9}{16}$
Since boiling point of pure ethanol = 360 K (given), so freezing point of pure ethanol = $T_{f} = \frac{4}{9} \times T_{b} = \frac{4}{9} \times 360 = 160 K$
So, $△ T_{f} = T_{f}^{0} - T_{f}^{S} = 160 - 152 = 8 K$ $(F r e e z i n g p o i n t o f s o l u t i o n (T_{f}^{S}) = 152 K, g i v e n)$
So, $\frac{△ T_{b}}{△ T_{f}} = \frac{k_{b} m}{k_{f} m} = \frac{9}{16} \Rightarrow △ T_{b} = \frac{9}{16} \times 8 = 4.5 K$
So, $T_{b}^{S} = T_{b}^{0} + △ T_{b} = 360 + 4.5 = 364.5 K$

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Similar questions

Q. The ratio of the boiling point (in K) of ethanol to its freezing point (in K) is 9:4, while the ratio of its latent heat of vaporization to its latent heat of fusion is 9:1. If the boiling point of pure ethanol is 360 K and freezing point of a solution of a compound P in ethanol is 152 K, then what is boiling point of same solution?

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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

$(K_{f}^{w a t e r}) = 1.86 K k g m o l^{- 1}$

Freezing point depression constant of ethanol

$(K_{b}^{w a t e r}) = 2.0 K k g m o l^{- 1}$

Boiling point elevation constant of water

$(K_{b}^{w a t e r}) = 0.52 K k g m o l^{- 1}$

Boiling point elevation constant of ethanol

$(K_{b}^{e t h a n o l}) = 1.2 K k g m o l^{- 1}$

Standard freezing point of water = 273 K

Standard freezing point of ethanol = 155.7 K

Standard boiling point of water = 373 K

Standard boiling point of ethanol = 351.5K

Vapour pressure of pure water = 32.8 mm Hg

Vapour pressure of pure ethanol = 40 mm Hg

Molecular weight of water $= 18 g m o l^{- 1}$

Molecular weight of ethanol $= 46 g m o l^{- 1}$

In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative.

The vapour pressure of the solution M is

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