Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
the mole fraction of constituent A in vapour in equilibrium with binary liquid solution A and B is 3/4. if the ratio of vapour pressure of A and B at the same temperature is 3:1, what is the molefraction of A in liquid phase?
a)1/2
b)1/4
c)1
d)3/4
from Raoult''s law, the partial vapour pressure of each component in the solutionis directly proportional to its mole fraction.
so partial pressure of A= P0(A) * XA = (3/4)P (let) (as mole fraction is directly prop. to partial pressure)
similarly, partial pressure of B= P0(B) * XB = (1/4)P
putting P0(A) =3 and P0(B) = 1 we get XA = XB =1/2 (A)
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !