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Grade 12th passPhysical Chemistry

The vapour pressure of water at 20C is 17.5 mm Hg. If 18 g of glucose is added to 178.2 g of H2O at 20C, the vapour pressure of soltion is? C is centrigate

Profile image of Saunak Nandi
8 Years agoGrade 12th pass
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2 Answers

Profile image of Arun
8 Years ago

by raults law 
P= Pa*Xa + Pb*Xb 
here Pa=O coz sugar is non volatile hence P=Pb*Xb 
Xb= 150/18 divided by 150/18 + 15/342 
Xb= 0.99 
PS= 0.99*17.5 = 17.4
Profile image of ankit singh
5 Years ago

No. of moles of glucose=18018=0.1

 
No. of moles of water=18178.2=9.9
 
Mole fraction of glucose=4.9+0.10.1=100.1=0.01
 
Using Raoult's law, as solute is non-volatile,
 
xβ=PAoPAoPA
 
PAo=vapour pressure of pure water
 
PA=vapour pressure of the solution
 
xβ=mole fraction of glucose
 
0.01=17.517.5PAPA=17.50.175=17.325
 
The vapor pressure of the mixture=17.325 mm Hg.