Home
Class 12
CHEMISTRY
Elevation in the boiling point for 1 mol...

Elevation in the boiling point for 1 molal solution of glucose is 2K. The depression in the freezing point for 2 molal solution of glucose in the same solvent is 2K. The relation between `K_(b) and K_(f)` is

A

`K_b=0.5K_f`

B

`K_b=2K_f`

C

`K_b=1.5K_f`

D

`K_b=K_f`

Text Solution

Verified by Experts

The correct Answer is:
B
Promotional Banner

Similar Questions

Explore conceptually related problems

The freezing point of 0.05 m solution of glucose in water is (K1 = 1.86°C m^(-1) )

Calculate elevation in boiling point for 2 molal aqueous solution of glucose. (Given K_b(H_(2)O) = 0.5 kg mol^(-1) )

For [CrCl_(3).xNH_(3)] , elevation in boiling point of one molal solution is double of one molal solution of glucose , hence x is if complex is 100% ionised :

The freezing point of a 0.05 molal solution of a non-electrolyte in water is: ( K_(f) = 1.86 "molality"^(-1) )

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

Elevation in boiling point of a solution of non-electrolyte in CCl_(4) is 0.60^(@)C . What is the depression in freezing point for the same solution? K_(f)(CCl_(4)) = 30.00 K kg mol^(-1), k_(b)(CCl_(4)) = 5.02 k kg mol^(-1)

An aqueous solution of 0.1 molal concentration of sucrose should have freezing point (K_(f)=1.86K mol^(-1)kg)

The rise in boiling point of a solution containing 1.8g glucose in 100g of a solvent is 0.1^(@)C . The molal elevation constant of the liquid is-

The molal depression constant for water is 1.86^(@)C . The freezing point of a 0.05-molal solution of a non-electrolyte in water is

Find the elevation in boiling point and (ii) depression in freezing point of a solution containing 0.520g glucose (C_(6)H_(12)O_(6)) dissolved in 80.2g of water. For water K_(f)= 1.86k//m. K_(b)=0.52 K//m.