Home
Class 12
CHEMISTRY
The relative decrease in VP of an aqueou...

The relative decrease in `VP` of an aqueous glucose dilute solution is found to be `0.018`. Hence, the elevation in boiling point is (it is given `1 molal` aqueous urea solution boils at `100.54^@)C` at `1 atm` pressure)

A

`0.018^(@)`

B

`0.18^(@)`

C

`0.54^(@)`

D

`0.03^(@)`

Text Solution

Verified by Experts

The correct Answer is:
C

`chi_(B)=(DeltaP)/P_(A)^(@)=0.018`
`DeltaT_(b)=K_(b).m rArr 0.54=K_(b) = xx1=K_(b)=0.54`
`m=chi_(B)/(1-chi_(B))xx1000/(Mw_(A))`
`m=0.018/0.982xx1000/18=1.0 rArr DeltaT_(b) = K_(b) xx m=0.54^(@)C`
Promotional Banner

Topper's Solved these Questions

  • SOLUTIONS

    CENGAGE CHEMISTRY|Exercise Ex 2.3 (Objective)|9 Videos
  • SOLUTIONS

    CENGAGE CHEMISTRY|Exercise Ex 2.1 (Objective)|10 Videos
  • SOLID STATE

    CENGAGE CHEMISTRY|Exercise Ex 1.2 (Objective)|9 Videos
  • SURFACE CHEMISTRY

    CENGAGE CHEMISTRY|Exercise Archives Subjective|2 Videos

Similar Questions

Explore conceptually related problems

Elevation of boiling point of 1 molar aqueous glucose solution ("density"=1.2g//ml) is

An aqueous solution of 0.01 M KCl cause the same elevation in boiling point as an aqueous solution of urea. The concetration of urea solution is :

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

What is the boiling point of 1 molal aqueous solution of NaCl (K_b)=0.52 K molal^-1]

Calculate elevation in boiling point for 2 molal aqueous solution of glucose (Given: K_(b(H_2O) =0.5K kg mol^(-1)))

The elevation in boiling points of 0.1m aqueous solutions of NaCI, CuSO_(4) and Na_(2)SO_(4) are in the ratio

Calculate relative lowering of vapour pressure of 0.161 molal aqueous solution