262even i have a similar doubt .
plz help.
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
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6 Answers
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.
