Vapour pressure of an equimolar mixture of benzene and toluene at a given temperature was found to be 80 mm Hg. If vapour above the liquid phase is condensed in a beaker, vapour pressure of this condensate at the same temperature was found to be 100 mm Hg. If the pure state vapour pressure of benzene and toluene is respectively x and y. Then determine value of x+2y40
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answer is 5.
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Detailed Solution
x+y=160yB=x160=xBlyT=y160=xTl100=xx160+yy160100160=x2+y2 100160=x2+160−x2 160100=x2+1602+x2−320x2x2−320x+16060=0x2−160x+8060=0x=160±1602−480602x=160±160160−1202x=160±4×2×102=160±802X=40, 120 (X=40 and Y=120 is rejected because benzene is more volatile then toluene) Y = 120, 40x+2y40=120+8040=20040=5