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Standard XIII

Chemistry

Depression in Freezing Point

Pure water fr...

Question

Pure water freezes at $273 K$ and $1 b a r$ . 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:

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Solution

The correct option is C
$Δ T_{f} = K_{f} \times m$

$= - 2 \times \frac{34.5 \times 2}{46}$

$= 2 \times 1.5$

$= 3$

Freezing point of ethanol +water mixture = 273 - 3 = 270 K

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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 kg mol}^{- 1}

. The figures shown below represent plots of vapour pressure (V.P.) versus temperature

(T)

. (Molecular weight of ethanol is

46 {g mol}^{- 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

Q. An aqueous solution freezes at 272.4 K while pure water freezes at 273 K. Determine the depression in freezing point of solution .

[Given :

K_{f} = 1.86 K k g m o l^{- 1}, K_{b} = 0.512 K k g m o l^{- 1}

and vapour pressure of water at 298 K = 23.756 mm of Hg].

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The correct option is C ΔTf=Kf×m =−2×34.5×246 =2×1.5 =3 Freezing point of ethanol +water mixture = 273 - 3 = 270 K

The correct option is C
$Δ T_{f} = K_{f} \times m$

$= - 2 \times \frac{34.5 \times 2}{46}$

$= 2 \times 1.5$

$= 3$

Freezing point of ethanol +water mixture = 273 - 3 = 270 K