wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The vapour pressure of an aqueous solution is found to be 750 torr at certain temperature T. If T is the temperature at which pure water boils under atmospheric pressure and same solution show elevation in boiling point ΔTb=1.04 K, find the atmospheric pressure (Kb=0.52 k kg mol1)

A
777
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
777.0
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
777.00
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

Elevation in boiling point ΔTb=kb×m
Where,m=molality
On substituting given values,
we get m=ΔTbkb=1.040.52
Molality (m)=2 mol/kg
So, Moles of solute (nsolute)=2 mol
Moles of solvent (nH2O)=100018=55.55 mol
The vapour pressure of the aqueous solution is given as 750 torr
According to Raoult'slaw:
PPsolutionPsolution=nsolutensolvent
P750750=2100018
P=750×2100018+750
P=27+750
P=777 torr

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Elevation in Boiling Point
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon