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प्रश्न
A mixture of benzene and toluene contains 30% by mass of toluene. At 30 °C, vapour pressure of pure toluene is 36.7 mm Hg and that of pure benzene is 118.2 mm Hg. Assuming that the two liquids form ideal solutions, calculate the total pressure and partial pressure of each constituent above the solution at 30 °C.
उत्तर
Given:
Mass percentage of toluene = 30 %W/W
Vapour pressure of liquid toluene `("P"_1^0)` = 36.7 mm Hg
Vapour pressure of liquid benzene `("P"_2^0)` = 118.2 mm Hg
To find:
1. Partial pressures of each constituent
2. Total pressure
Formulae:
1. `"P"_1 = "P"_1^0 "x"_1 and "P"_2 = "P"_2^0 "x"_2`,
2. P = P1 + P2
Calculation:
Molar mass of toluene, C7H8 = (7 ×12) + (8 × 1) = 92 g mol-1
Molar mass of benzene, C6H6 = (6 ×12) + (6 × 1) = 78 g mol-1
Now, 30 % W/W toluene means 30 g toluene in 100 g solution.
Thus, mass of benzene = 100 – 30 = 70 g
Number of moles of toluene, C7H8 = `"n"_"A" = (30 "g")/(92 "mol")` = 0.326 mol
Number of moles of benzene, C6H6 = `"n"_"B" ="70 g"/(78 "mol")` = 0.897 mol
Total number of moles `("n"_"A" + "n"_"B") = 0.326 "mol" + 0.897 "mol"` = 1.223 mol
Mole fraction of toluene `("x"_("C"_7"H"_6)) = "n"_"A"/("n"_"A" + "n"_"B") = 0.326/1.223 = 0.2666`
Mole fraction of benzene `("x"_("C"_6"H"_6))` = 1.0 - 0.2666 = 0.7334
Now, using formula (i),
`"P"_("C"_7 "H"_8) = "P"_("C"_7"H"_8)^0 xx "x"_("C"_7"H"_8)` = 36.7 mm Hg × 0.2666 = 9.78 mm Hg
`"P"_("C"_6 "H"_6) = "P"_("C"_6"H"_6)^0 xx "x"_("C"_6"H"_6)` = 118.2 mm Hg × 0.7334 = 86.7 mm Hg
Now, using formula (ii),
Vapour pressure of the solution,
P = `"P"_("C"_7 "H"_8) + "P"_("C"_6 "H"_6) = 9.78 + 86.7 = 96.48 = 96.5` mm Hg
∴ Partial pressures of toluene and benzene are 9.78 mm Hg and 86.7 mm Hg, respectively.
∴ Total pressure above the solution is 96.5 mm Hg.
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