You visited us 1 times! Enjoying our articles? Unlock Full Access!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

15.4 g mol

No worries! We‘ve got your back. Try BYJU‘S free classes today!

17.3 g mol^-1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

20.4 g mol

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C
17.3 g mol^-1

First boiling point of water = 100⁰C

Final boiling point of water = 100.52⁰

Suggest Corrections

Similar questions

3 g

200 m L

(100^{\circ} C)

{100.52}^{\circ} C

K_{b}

0.6 K / m

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

{100.34}^{\circ} C

K_{b}

= {0.52}^{\circ} C / m

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

48

1200

K_{b}

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

= 100

0.1 m

^{\circ} C

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

= 100^{\circ} C

Join BYJU'S Learning Program