Question

The mass of glucose that should be dissolved in 50g of water in order

Answer 1

Answer:Solution:

P−Ps /p=w1MM/w2M1

To produce same lowering of vapour

pressure,P−Ps/P will be same for both cases.

So,  W(Glucose)×18/50×180=W(urea)×/50×60

W(Glucose)= weight of glucose

W(Glucose) = weight of urea

or W(Glucose)×18/50×180=1×18/50×60

W(Glucose) = 3