at 300K vapour pres. of pure benzene and pure tolune are 100mm and 30mm of Hg rep. also they form ideal soln.a soln. of benzene and tol. is prep. by mixin 3 mole of tolun in 2 moles of benz. at 300K.
1)at wht external pressure at 300k the lst drop of given soln. will disappear. answer= 125/3 mm hg
2) if vapours which is in equilibrium with soln. are condensed , mole frac. of benzene in first drop of liquid formed will be? ans=20/29
71Okay. We calculate some general things first.
PTotal = 58 mm Hg
XA = 3/5 , XB = 2/5
YA = 9/29 , YB = 20/29
When last drop is remaining, we need to find the pressure exerted by vapour. In that case,
3/5 (which is the mole fraction of the original solution is now the mole fraction of the vapour because almost all has evaporated) = P°Axa(P°Axa + (1-xa)P°B)
We get xa = 5/6 , xb = 1/6
Hence, Pressure required = 5/6 * 30+ 1/6 * 100 = 125/3 mm Hg
A = Toulene , B = Benzene
21thanks vivek for soln.
aditya utilize partial pressure is proportional to mole fraction.
Ybenzene=pbenzene/ptotal
We can get Pbenzene and ptotal from Raults law at equilibrium.
