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Standard XII

Chemistry

Abnormal Colligative Properties

Pure water fr...

Question

Pure water freezes at 273 K and 1 bar. The addition of 34.5g of ethanol to 500g of water changes the freezing point of the solution. Use the freezing point depression constant of water as 2K kg $m o l^{- 1}$ . The figures shown below represent plots of vapour pressure (V.P.) versus temperature (T). [Molecular weight of ethanol is 46g $m o l^{- 1}$ ]

Among the following, the option representing change in the freezing point is

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Solution

Step 1:
The number of moles of ethanol n = $\frac{34.5 g}{46 g {m o l}^{- 1}} = 0.75 m o l$
Molality m = $\frac{0.75 m o l}{0.5 K g} = 1.5 m$
Step 2:
The depression in freezing point $Δ T_{f} = K_{f}^{m} = 2 K K g {m o l}^{- 1} \times 1.5 m = 3 K$
The freezing point of pure water is 273 K.
Freezing point of solution = 273 − 3 = 270 K.

Option (A) represents the correct graph. In this graph, the freezing point of the solution is shown as 270 K. Also with the increase in temperature, the vapour pressure of pure water and mixture increases.
Hence, option (A) is correct.

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Similar questions

Q. Pure water freezes at

273 K

and

1

bar. The addition of

34.5 g

of ethanol to

500 g

of water changes the freezing point of the solution. Use the freezing point depression constant of water as

2 K k g m o l^{- 1}

. The figures shown below represent plots of vapour pressure (V.P.) versus temperature (T). [molecular weight of ethanol is

46 g m o l^{- 1}

]
Among the following, the option representing change in the freezing point is

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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