MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

⇐ BackProceed to Checkout ⇒
There are no items in this cart.
Continue Shopping
Menu
Grade: 12 
        The relative lowering of vapour pressure of an aqueous solution containing non volatile solute is 0.0125. the molality of solution is
one year ago

Answers : (1)

Arun
18691 Points 
							
Dear student
 
Relative lowering of vapour pressure = mole fraction of solute (Roult's law)
p- p(s)/p = x2
 
[p - p(s)]/p = w *M/W*m = 0.0125
 
w/m *W = 0.0125/18 = 0.00070
 
Molality = 0.00070 * 1000
= 0.70 M
 
Regards
Arun (askIITians forum expert)
one year ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

Other Related Questions on Physical Chemistry

if a mixture of oxygen and ozone is having the rms velocity = 12.92 m/s at 300K. Calculate mole... Answer & Earn Cool Goodies
if a mixture of oxygen and ozone is having the rms velocity = 12.92m/s at 300K. Calculate the mole... Answer & Earn Cool Goodies
A vessel of 120 ml capacity current contains a certain mass of gas at 20 degree Celsius and 750 mm...
What is the unit of first order reaction,second order reaction,third order reaction and formula of...
Hello sir I got an dout about static friction Static friction can not cause an acceleration or not
View all Questions »


Course Features

  • 731 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 141 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers To Win!!! Click Here for details