Answer to Question #139174 in General Chemistry for Janesel

Question

Assuming Raoult’s law applies, calculate the vapor pressure of a solution of 45 g of glucose (C6H12O6) in 95g of water at 25 degrees Celsius

Answer 1

According to Raoult's Law

(P°-P)/P°= x2 where

P°= Vapour Pressure of water .

P= Vapour Pressure of Solution .

x2= mole-fraction if the Solute

=x2= n2/(n1+n2) where

n1= no.of moles of Solvent=95/18=5.277

n2= no.of moles of Solute=45/180=0.25 So n1+ n2= 5.277+ 0.25= 5.5275 .

n2/( n1+n2)= 0.25/5.527= 0.0452

Now (P°-P)/P°= 0.0452

vapour pressure of the pure water is 23.7695 at 25 degrees Celsius.

So P°-P= 23.7695 x 0.0452= 1.074

So P=Vapour Pressure of Solution

= 23.7695- 1.074 = 22.6955mm.