A solution containing benzene and toluene in the mole ratio of 6:4 has a vapour pressure of 667.06 mm of Hg at $85^{o} C$ .A solution at the same temperature containing benzene and toluene in the ratio of 4:6 has a vapour pressure of 559.74. Calculate the mole ratio of benzene-toluene which would boil at this temperature.Atmospheric pressure =760 mm of Hg.
if ratio is x:y then x+y is

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Solution

$χ_{B} = \frac{6}{10} = 0.6, χ_{T} = 0.4$
$667.06 = P = P_{B}^{o} \times 0.6 + P_{T}^{o} \times 0.4 . . . (i)$
$559.74 = P = P_{B}^{o} \times 0.4 + P_{T}^{o} \times 0.6 . . . (i i)$
$107.32 = 0.2 (P_{B}^{o} - P_{T}^{o})$
$P_{B}^{o} - P_{T}^{o} = 536.6 . . . (i i i)$
$P_{B}^{o} = 536.6 + P_{T}^{o}$
Putting the value of $P_{B}^{o}$ in equation (i), we get
$667.06 = (536.6 + P_{T}^{o}) 0.6 + P_{T}^{o} \times 0.4$
$P_{T}^{o} = 345.1$
Similarly, $P_{B}^{o} = 881.7$
$760 = 881.7 χ_{B} = 345.1 (1 - χ_{B})$
Solve, $χ_{B} = 0.77, χ_{T} = 0.23$
Ratio=0.77:0.23=3:1
So x+y = 3+1 =4

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