Home
Class 11
CHEMISTRY
Total vapour pressure of mixture of 1 m...

Total vapour pressure of mixture of 1 mole of volaile components A (`P_(a^(%)`)=100 mm Hg) and 3 mole of volatile component B(`P_B^(@) =80 mm Hg`) is 90 mm Hg. For such case:

A

There is positive deviation from Rsoult's law

B

boiling point has been lowered

C

force of attraction between A and B is weaker than that between A and A or betweenB and B

D

All the above statement are correct

Text Solution

Verified by Experts

The correct Answer is:
d
Promotional Banner

Topper's Solved these Questions

  • DILUTE SOLUTION

    NARENDRA AWASTHI|Exercise Level 1 (Q.62 To Q.91)|1 Videos

  • DILUTE SOLUTION

    NARENDRA AWASTHI|Exercise Level 2|3 Videos

  • CHEMICAL EQUILIBRIUM

    NARENDRA AWASTHI|Exercise Level 2|1 Videos

  • ELECTROCHEMISTRY

    NARENDRA AWASTHI|Exercise Level 3 - Subjective Problems|1 Videos

Similar Questions

Explore conceptually related problems

Total vapour pressure of mixture of 1 mole of volatile component A (P_(A)^(@)=100 mm Hg) and 3 mole of volatile component B (P_(B)^(@)=80 mm Hg) is 90 mm Hg . For such case :

Total vapour pressure of mixture of 1 mol of volatile component A( p_(A)^(@)=100 mm Hg) and 3 mol of volatile component B( p_(B)^(@) =60 mm Hg) is 75 mm. For such case:

Total vapour pressure of mixture of 1 mol of volatile component A(P_(A^(@))=100mmHg) and 3 mol of volatile component B(P_(B^(@)=60 mmHg) is 75mm . For such case:

Total vapour pressure of mixture of 1 mol volatile component A(P^(@)A="100 mm Hg") and 3 mol of volatile component B(P^(@)B="60 mm Hg") is 75 mm. For such case -

A mixture contains 1 mole of volatile liquid A (P_(A)^(@) =100mm Hg) and 3 moles of volatile liquid B (P_(B)^(@) =80 mm Hg). If solution behaves ideally, the total vapour pressure of the distillate is :

A mixture contains 1 mole of volatile liquid A ("P"_("A")^(0)=80"mm Hg"). and 3 moles of volatile liquid B ("P"_("A")^(0)=80"mm Hg"). If solution behaves ideally, the total vapour pressure of the distillate is