CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
Question

At 300 K, the vapour pressure of an ideal solution containing 1 mole of A and 3 moles of B is 500 mm Hg. At the same temperature, 2 moles of B are added to this solution. The vapour pressure of the solution increases by 10% of the original vapour pressure. The correct statement(s) about the vapour pressure are :

A
The vapour pressure of A in the pure state is 1300 mm Hg
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
The vapour pressure of B in the pure state is 650 mm Hg
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
The ratio of the final vapour pressure to the initial vapour pressure is 2 : 1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
The ratio of the vapour pressure of pure B to the vapour pressure of pure A is 13 : 1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct options are
B The vapour pressure of B in the pure state is 650 mm Hg
D The ratio of the vapour pressure of pure B to the vapour pressure of pure A is 13 : 1
nA = 1; nB = 3; PT = 500 mm Hg
When 2 moles of B is added
nA = 1; nB = 5
As it is given that vapour pressure increases by 10%,
PT=500+500×10100=550
500 = PA(11+3)+PB(31+3)
2000 = PA+3PB(1)
550 = PA(16)+PB(56)
3300 = PA+5PB(2)
Solving (1) and (2) we get,
PA=50 mm Hg PB=650 mm Hg
PAPB=65050=13:1

flag
Suggest Corrections
thumbs-up
1
BNAT
mid-banner-image
similar_icon
Similar questions
View More