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Standard XII

Chemistry

Elevation in Boiling Point

To observe an...

Question

To observe an elevation of boiling point of ${0.05}^{o} C$ , the amount of a solute $(m o l . w t . = 100)$ to be added to $100 g$ of water $(K_{b} = 0.5)$ is:

2 g

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0.5 g

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1 g

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0.75 g

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Solution

The correct option is C $1 g$
Elevation of boiling point,

$Δ T_{b} = \frac{w \times K_{b} \times 1000}{M \times W}$

(Here, $w$ and $W =$ weights of solute and solvent respectively, $M =$ molecular weight of solute and $K_{b} =$ constant).

On substituting values, we get

$0.05 = \frac{w \times 0.5 \times 1000}{100 \times 100}$

or $w = \frac{0.05 \times 100 \times 100}{0.5 \times 1000} = 1 g$

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Similar questions

Q. Assertion :When methanol is added to water, boiling point of water increases. Reason: When a volatile solute is added to a volatile solvent, elevation in boiling point is observed.

Assetion (A) When methyl alcohol is added to water, boiling point of water increases.
Reason (R) When a volatile solute is added to a volatile solvent elevation in boiling point is observed.

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

]

The boiling point of water at 735 torr is 99.07°C.The mass of NaCl added in 100 g water to make its boiling point 100°C is ( K_b = 0.51 K kg mol^-1)

ans: 5.34 g

how the answer came

I want step by step detailed explanation

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

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To observe an elevation of boiling point of 0.05oC, the amount of a solute (mol.wt.=100) to be added to 100g of water (Kb=0.5) is:

The correct option is C 1 gElevation of boiling point,ΔTb=w×Kb×1000M×W(Here, w and W= weights of solute and solvent respectively, M= molecular weight of solute and Kb= constant).On substituting values, we get0.05=w×0.5×1000100×100or w=0.05×100×1000.5×1000=1 g

To observe an elevation of boiling point of ${0.05}^{o} C$ , the amount of a solute $(m o l . w t . = 100)$ to be added to $100 g$ of water $(K_{b} = 0.5)$ is: