Calculate vapor pressure of an aqueous solution of 1 molal glucose solution at 100 degree Celsius.....

Given:

An aqueous solution of 1 molal glucose solution at 100 ºC

We need to find the vapor pressure

Solution

We know that

Molarity =W2/Mw2×W1×1000 {W2 = mass of solute in grams, W1 = mass of solvent in grams}
1.0=W2×1000Mw2×W1
We have studied that

K/W1=K’=1.0/1000=0.001
On applying Raoult’s law for a dilute solution, we get
P−PSP=W2/Mw2×W1×Mw1 {Mw1=18}
760−PS760=0.001×18
PS=760−13.68
PS=746.32mmHg

Hencevapor pressure =PS=746.32mmHg

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