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Standard XII

Chemistry

Elevation in Boiling Point

The molal ele...

Question

The molal elevation constant of water is $0.51$ . The boiling point of $0.1$ molal aqueous NaCl solution is nearly:

{100.05}^{o}

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{100.1}^{o}

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{100.2}^{o}

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{101.0}^{o}

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Solution

The correct option is B ${100.05}^{o}$ C
Let us consider the problem:
$Δ T_{b} = K_{b} . m = 0.51 \times 0.1 = {0.051}^{\circ}$
Now b.p of solution $= 100 + Δ T_{b}$
Implies that
$= 100 + 0.051$
Hence,
$= {100.051}^{\circ} C$

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Similar questions

Q. The molal elevation constant of water is

{0.51}^{\circ} C

K g m o l^{- 1}

. The boiling point of 0.1 molal aqueous NaCl solution is nearly :

Which of the following aqueous solutions will have the boiling point

{102.2}^{o} C

The molal elevation constant for water is

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

The molal elevation constant of water = 0.52⁰C. The boiling point of 1.0 molal aqueous KCl solution (assuming complete dissociation of KCl), therefore, should be

Q. The molality of an aqueous dilute solution containing non-volatile solute is

0.1 m

. What is the boiling temperature (in

^{\circ} C

) of solution? (Boiling point elevation constant,

K_{b} = 0.52 k g m o l^{- 1} K

; boiling temperature of water

= 100^{\circ} C

A B_{2}

10 %

dissociated in water to

A^{2 +}

and

B^{-}

. The boiling point of a

10.0

molal aqueous solution of

A B_{2}

^{\circ} C .

(Round off to the nearest Integer).
[Given: Molal elevation constant of water

K_{b} = 0.5 {K kg mol}^{- 1}

, boiling point of pure water

= 100^{\circ} C

]

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The molal elevation constant of water is 0.51. The boiling point of 0.1 molal aqueous NaCl solution is nearly:

The correct option is B 100.05oCLet us consider the problem:ΔTb=Kb.m=0.51×0.1=0.051∘Now b.p of solution =100+ΔTbImplies that=100+0.051Hence,=100.051∘C

The molal elevation constant of water is $0.51$ . The boiling point of $0.1$ molal aqueous NaCl solution is nearly:

The correct option is B ${100.05}^{o}$ C
Let us consider the problem:
$Δ T_{b} = K_{b} . m = 0.51 \times 0.1 = {0.051}^{\circ}$
Now b.p of solution $= 100 + Δ T_{b}$
Implies that
$= 100 + 0.051$
Hence,
$= {100.051}^{\circ} C$