Request a call back

Join NOW to get access to exclusive study material for best results

CBSE Class 12-science Answered

2.175 gm of non volatile solute is added in 39 gm of organic liquid.The lowering in the vapour pressure is 40 mm oh Hg. Calculate molar weight of solute>molecular weight of organic liquid is 78.Assuming solution to be dilute vapour pressure of pure liquid is 640mm of Hg.
Asked by Asif Ansari | 26 Jun, 2011, 12:33: AM
answered-by-expert Expert Answer
Mass of non volatile solute(WB) = 2.175 g
Mass of organic liquid (WA) =39 g
 Given lowering in the vapour pressure ( P0A-PA) = 40 mm of Hg
Vapour pressure of  pure liquid, P0A= 640 mm of Hg

molar mass of organic liquid = 78 g mol-1

For dilute solution,

(P0A-P/P0A  = (WB XMA) /(MB XWA)
   40/640 = (2.175 X78)/(MB X39)
MB =69.6 g mol-1
Therefore molar weight of solute =69.6 g mol-1
Answered by | 27 Jun, 2011, 04:29: PM
CBSE 12-science - Chemistry
Asked by sameerteli003 | 08 Apr, 2024, 11:48: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 12-science - Chemistry
Asked by rashmij34 | 27 Feb, 2024, 04:42: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 12-science - Chemistry
Asked by sagarmishra | 27 Feb, 2024, 04:01: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 12-science - Chemistry
Asked by kalandi.charan.407 | 08 Feb, 2024, 01:42: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 12-science - Chemistry
Asked by premkhare2006 | 24 Jan, 2024, 09:50: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 12-science - Chemistry
Asked by saritanohar22 | 13 Jan, 2024, 01:25: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
Get Latest Study Material for Academic year 24-25 Click here
×