14.1 m aq. Solution of a salt shows elevation in boiling point equal to 3 times of its kb. Determine (i) freezing point of solution (ii) osmotic pressure at 27^° c HINT: (i) 0- Tf'= 3×1.86×1 (ii) pi=3×1×0.0821×300atm.

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An aqueous solution freezes at -0.2°C. What is the molality of the solution. Determine also the-

(I) elevation in boiling point

(II) Lowering of vapour pressure

Given Kf=1.86°C Kg/mol Kb=0.512°C Kg/mol.

Vapour pressure of water=23.756 mm

Q. A solution of urea in water has a boiling point of

{100.18}^{\circ} C .

Calculate the freezing point of the same solution. Molal depression and elevation constants for water

K_{f}

and

K_{b}

are

{1.86}^{\circ} C k g m o l^{- 1}

and

{0.512}^{\circ} C k g m o l^{- 1}

respectively.

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

[Given :

K_{f} = 1.86

K kg mol

^{- 1}

K_{b} = 0.512

K kg mol

^{- 1}

and vapour pressure of water at

298 K = 23.756

mm of Hg].

Question 49
Match the laws given in Column I with expressions given in Column II.
$\begin{matrix} C o l u m n I & C o l u m n I I A . R a o u l t^{'} s l a w & 1. Δ T_{f} = K_{f} m B . H e n r y^{'} s l a w & 2. π = C R T C . E l e v a t i o n o f b o i l i n g p o i n t & 3. p = x_{1} p_{1}^{0} + x_{2} p_{2}^{0} D . D e p r e s s i o n i n f r e e z i n g p o i n t & 4. Δ T_{b} = K_{b} m E . O s m o t i c p r e s s u r e & 5. p = K_{H} . x \end{matrix}$

Q. For an aqueous solution, freezing point is

- {0.186}^{0} C

. Elevation of the boiling point is the same solution is : (

K_{f} = {1.86}^{0} m o l^{- 1} k g

and

K_{b} = {0.512}^{0} m o l^{- 1} k g

)

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