Question

What will be the vapor pressure of the solution of 18gm glucose dissloved in 178.2 grams of water at?

Answer 1

Molecular mass of water =2×1+1×16=18g=2×1+1×16=18gFor 178.2g178.2g water nA=9.9nA=9.9Molecular mass of glucose =6×12+12×1+6×16=180g=6×12+12×1+6×16=180gFor 18g18g glucose nB=0.1nB=0.1XB=0.10.1+9.9)XB=0.10.1+9.9)=0.01=0.01XA=0.99XA=0.99For lowering of vapour pressure ,p=p0AXA=P0A(1−XB)p=pA0XA=PA0(1−XB)P=760(1−0.01)P=760(1−0.01)=760−7.6=760−7.6=752.4torr