Question

A mixture of two immiscible liquids nitrobenzene & water boiling at 99^^oC has a partial vapour pressure of water 733 mm & that of nitrobenzene 27 mm. The ratio of the weights of nitrobenzene to the water in the distillate is

ans 4:1 plz explain aboout it

Answer 1

At a boiling temperature of 99 oC, number of moles of two liquids nitrobenzene and water are proportional to their vapour pressures.

So, nnitrobenznenwater=P0nitrobenzeneP0water

Or in other words,

PonitrobenzenePowater=Wnitrobenzene×MwaterWwater×Mnitrobenzeneor WnitrobenzeneWwater= Ponitrobenzene×MnitrobenzenePowater×Mwater=27×123733×18=0.25or WnitrobenzeneWwater=0.25=14

Answer 2

Answer:

4 :1

Explanation:

pw = 733mm ( 1 atm = 760)

pn = 27mm

pW = p°w ×xw

xw = 733/760

0.96

as

xn = 0.036

0.964/0.036

= w × 123 / 18 × wn

xw/xn = n /nN = 3.8999

approx 4

pls mark as brainliest answer