JEE Main · 2010 · 11 Apr 2010 · Offline · Solutions

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On mixing, heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components (heptane and octane) are 105 kPa and 45 kPa respectively. Vapour pressure of the solution obtained by mixing 25.0 g of heptane and 35 g of octane will be (molar mass of heptane

= 100 g mol^{- 1}

and of octane

= 114 g mol^{- 1}

(A)72.0 kPa✓ Correct
(B)36.1 kPa
(C)96.2 kPa
(D)144.5 kPa

Answer: (A)

Solution

First find the mole fractions. Heptane gives

\frac{25}{100} = 0.25

mol and octane

\frac{35}{114} = 0.307

mol, a total of

0.557

mol, so

x_{hep} = 0.45

and

x_{oct} = 0.55

. For an ideal solution Raoult's law makes the total pressure the sum of the two contributions,

P = x_{hep} P_{hep}^{\circ} + x_{oct} P_{oct}^{\circ} = 0.45 \times 105 + 0.55 \times 45 = 47.25 + 24.75 = 72.0

kPa.

JEE Main · 2010 · 11 Apr 2010 · Offline · Single correct option

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JEE Main · 2010 · 11 Apr 2010 · Offline · Solutions

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