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).

Question

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 (B) 36.1 kPa (C) 96.2 kPa (D) 144.5 kPa

Tardigrade Admin · Accepted Answer

Mole fraction of Heptane = (25 /100/(25)100+(35)114 = (0.25/0.557) = 0.45 X/Hep tane) = 0.45 . ∴ Mole fraction of octane = 0.55 = Xoctane Total pressure = ∑ XiPi0 = (105 × 0.45) + (45 × 0.55) kPa = 72.0 KPa

$72.0 k P a$

$36.1 k P a$

$96.2 k P a$

$144.5 k P a$

Mole fraction of Heptane $= \frac{25/100}{\frac{25}{100} + \frac{35}{114}} = \frac{0.25}{0.557} = 0.45$
$X_{He p t an e} = 0.45.$
$∴$ Mole fraction of octane $= 0.55 = X_{oc t an e}$
Total pressure $= \sum X_{i} P_{i}^{0}$
$= (105 \times 0.45) + (45 \times 0.55) k P_{a}$
$= 72.0 K P a$

72.0kPa

36.1kPa

96.2kPa

144.5kPa

Mole fraction of Heptane =10025​+11435​25/100​=0.5570.25​=0.45 XHeptane​=0.45. ∴ Mole fraction of octane =0.55=Xoctane​ Total pressure =∑Xi​Pi0​ =(105×0.45)+(45×0.55)kPa​ =72.0KPa

$72.0 k P a$

$36.1 k P a$

$96.2 k P a$

$144.5 k P a$

Mole fraction of Heptane $= \frac{25/100}{\frac{25}{100} + \frac{35}{114}} = \frac{0.25}{0.557} = 0.45$
$X_{He p t an e} = 0.45.$
$∴$ Mole fraction of octane $= 0.55 = X_{oc t an e}$
Total pressure $= \sum X_{i} P_{i}^{0}$
$= (105 \times 0.45) + (45 \times 0.55) k P_{a}$
$= 72.0 K P a$