An organic liquid A (immiscible with water) when boiled together with water, the boiling point is 90^(@)C at which the partial vapour pressure of water is 526 mm Hg. The atmospheric pressure is 736 mm Hg. The weight ratio of the liquid and water collected is 2.5 : 1. Calculate the molecular weight of the liquid.

An organic liquid A (immiscible with water) when boiled together with

1.	An organic liquid A (immiscible with water) when boiled together with water, the boiling point is 90^(@)C at which the partial vapour pressure of water is 526 mm Hg. The atmospheric pressure is 736 mm Hg. The weight ratio of the liquid and water collected is 2.5 : 1. Calculate the molecular weight of the liquid.
Answer» <P> Solution :For two immiscible liquid, at the boiling point, ratio of the NUMBER of MOLES of the two liquids in the vapour phase is equal to the ratio of their vapour pressure, i.e., `(n_(l))/(n_(w))=(p_(l))/(p_(w))"...(i)"` GIVEN, at the boiling point, `p_(w)="526 mm and p"_("total")="736 mm"` `THEREFORE""p_(l)=7360526=210 mm` From eqn. (i), `(W_(l)//W_(l))/(W_(w)//M_(w))=(p_(l))/(p_(w)) or (W_(l))/(W_(w))=(p_(l)xxM_(l))/(p_(w)xxM_(w))""therefore""(2.5)/(1)=(210xxM_(l))/(526xx18) or M_(l)="112.7 g mol"^(-1)`