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Standard XII

Chemistry

Depression in Freezing Point

If the elevat...

Question

If the elevation in boiling point of a solution of non-volatile, non-electrolytic and non-associating solute in a solvent $(K_{b} = x K k g m o l^{- 1})$ is $y K$ , then the depression in freezing point of solution of same concentration would be:

$[K_{f}$ of the solvent $= z K k g m o l^{- 1}]$

\frac{2 x z}{y}

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\frac{y z}{x}

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\frac{x z}{y}

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\frac{y z}{2 x}

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\frac{x z}{2 y}

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Solution

The correct option is B $\frac{y z}{x}$
$Δ T_{b} = K_{b} \times m$

$∴ y = x \times m o l a l i t y$

Again, $Δ T_{f} = K_{f} \times m o l a l i t y = z \times m o l a l i t y$

$= z \times \frac{y}{x}$

$= \frac{y z}{x}$ .

Hence, option B is correct.

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Similar questions

Q. A solvent X freezes at

240^{\circ} C

. If

0.25

molal solution of a non volatile solute in pure X causes freezing point depression of

3^{\circ} C

, then

K_{f}

(in

{K kg mol}^{- 1}

) for solvent X is:

Assume complete association/dissociation of solute.

Q. Assertion :A non volatile solute is mixed in a solution then elevation in boiling point and depression in freezing point both are

2 K

. Reason: Elevation in boiling point and depression in freezing point both depend on melting point of non-volatile solute.

Q. For a dilute solution containing

2.5 g m

of a non-volatile non-electrolyte solute in

100 g m

of water, the elevation in boiling point at

1 a t m

pressure is

2^{o} C

. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of

H g

) of the solution is (

K_{b} = 0.76 K k g m o l^{- 1})

For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in $100 g$ of water, the elevation in boiling point at $1 a t m$ pressure is $2 ℃$ . Assuming that the concentration of solute is much lower than the concentration of solvent, the vapour pressure $(in mm of Hg)$ of the solution is:

[Take : $K_{b} = 0.76 K k g m o l^{- 1}$ ]

Q. Given that,

Δ T_{f}

is the depression in freezing point of the solvent in a solution of a non-volatile non-electrolytic solute of molality m, then quantity -

lim m \to 0 (Δ T_{f} / m)

is ___ (equal\different) to molal depression constant.

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If the elevation in boiling point of a solution of non-volatile, non-electrolytic and non-associating solute in a solvent (Kb=x K kg mol−1) is y K, then the depression in freezing point of solution of same concentration would be: [Kf of the solvent =z K kg mol−1]

The correct option is B yzxΔTb=Kb×m ∴y=x×molalityAgain, ΔTf=Kf×molality=z×molality =z×yx =yzx.Hence, option B is correct.

If the elevation in boiling point of a solution of non-volatile, non-electrolytic and non-associating solute in a solvent $(K_{b} = x K k g m o l^{- 1})$ is $y K$ , then the depression in freezing point of solution of same concentration would be:

$[K_{f}$ of the solvent $= z K k g m o l^{- 1}]$

The correct option is B $\frac{y z}{x}$
$Δ T_{b} = K_{b} \times m$

$∴ y = x \times m o l a l i t y$

Again, $Δ T_{f} = K_{f} \times m o l a l i t y = z \times m o l a l i t y$

$= z \times \frac{y}{x}$

$= \frac{y z}{x}$ .

Hence, option B is correct.