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Elevation in Boiling Point

The boiling p...

Question

The boiling point of water (100°C) becomes 100.52°C, if 3 grams of a non-volatile solute is dissolved in 200ml of water. The molecular weight of

solute is

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15.4 g mol

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17.3 g mol^-1

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20.4 g mol

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Solution

The correct option is C
17.3 g mol^-1

First boiling point of water = 100⁰C

Final boiling point of water = 100.52⁰

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Similar questions

The boiling point of water (100°C) becomes 100.52°C, if 3 grams of a non-volatile solute is dissolved in 200ml of water. The molecular weight of

solute is

3 g

200 m L

(100^{\circ} C)

{100.52}^{\circ} C

K_{b}

0.6 K / m

50 %

{100.52}^{o}

[K_{b} = 0.52

m o l^{- 1}

K_{f} = 1.86

m o l^{- 1}]

(M . w t : 60)

90 g

0.02

58 g {m o l}^{- 1}

48

1200

K_{b}

= 0.513 k g {m o l e}^{- 1}

= 100

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