Home
Class 12
CHEMISTRY
Figure explains elevation in boiling poi...

Figure explains elevation in boiling point when a non-volatile solute is added to a solvent. Variation of vapour pressure with temperaure and elevation in boiling point is marked.

Elevation in b.p of an aqueous urea solution is `0.52^(@) (K_(b) = 0.52^(@) mol^(-1)kg)`. Hence, mole fraction of urea in this solution is:

A

0.982

B

0.0567

C

0.943

D

0.018

Text Solution

Verified by Experts

The correct Answer is:
D
Promotional Banner

Similar Questions

Explore conceptually related problems

Elevation in b.p. of an aqueous urea solution is 0.52^(@), (K_(b)=0.622^(@) "mol"^(-1)kg) Hence, mole-fraction of urea in this solution is :

Figure explains elevation in boiling point when a non-volatile solute is added to a solvent. Variation of vapour pressure with temperaure and elevation in boiling point is marked. Given that DeltaT_(b) is the elevation in boiling point of the solvent in a solution of molarity m then lim_(m rarr 0) ((Delta T_(b))/(m)) is equal to:

Figure explains elevation in boiling point when a non-volatile solute is added to a solvent. Variation of vapour pressure with temperaure and elevation in boiling point is marked. Ratio of DeltaT_(b)//K_(b) of 6% AB_(2) and 9% A_(2)B (AB_(2) and A_(2)B both are non-electrolytes) is 1 mol/kg in both cases. Hence, atomic masses of A and B are respectively.

Elevation in boiling point of an aqueous solution of urea is 0.52 ( k_(b) "for water"=0.52K"molality"^(-1)) . The mole fraction is urea in this solution is :