Isotonic solution
πGlycerine=πGlucose
n1V1RT=n2V2RT
i.e,n1V1=n2V2
Where, n1 is the no. of moles of glycerine
V1 is volume of glycerine
n2 is the no. of moles of glucose
V2 is volume of glucose
WGlycerineMGlycerine×1V1=WGlucoseMGlucose×1V2
10.2MGlycerine×11000=2180×1100
MGlycerine=10.2×180×1002×1000
=91.8gm
Hence, the molecular mass of glycerine =91.8 gm