Question

18 g of glucose dissolved in 500 g of water .Find the depression of freezing point

Answer 1

We know depression of freezing point is a colligative property.

∆T=Kf*m(where Kf is cryoscopic constant,m is molality)

Kf of water is 1.86°C/m

m=No of moles/Mass of solvent in kg=(18/180)/(500/1000)=1/5

So, depression is 1.86*1/5=0.372°C

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