Calculate % of lowering of vapour pressure if 36 g glucose is added in 312g of benzene

Answer 1

It is assumed throughout the calculation that these two components when mixed make an ideal solution.

So from what I've learned,in an ideal solution of two components,the lowering of vapour pressure of one component equals to the mole fraction of the other component in aqueous zone.(From Raoult's law)

To get the percentage multiply it by 100℅,which will give you the required answer.

Correct me if I have mention anything which is not correct!

Answer 2

I believe that's the mole fraction of the other component, as stated, then times the vapor pressure of the benzene.
Percent lowering = [(vp lowering-Po)/Po]*100 = ?