Question

In glucose's aqueous solution, mole fraction of water is 0.997 . Then what will be solution's molality? How?

Answer 1

(XA x 1000)/ (XB x mol.wt. of B) = molality ‘m’

XB =1-XA

XB =mole fraction of water

XB=0.997

(XA x 1000)/ (1-XA)x mol.wt. of B = molality ‘m’

XA = mole fraction of glucose

XA =1-XB=1-0.997=0.003

m=0.003x 1000/(0.997 x 18)

m= 0.167 mole/kg