Heptane and octane form an ideal solution. At $373 K$ , the vapour pressures of the two liquid components are $105.2 k P a$ and $46.8 k P a$ respectively. What will be the vapour pressure of a mixture of $26.0 g$ of heptane and $35 g$ of octane?

Open in App

Solution

The molar masses of heptane and octane are $100$ g/mol and $114$ g/mol respectively.

$26$ g of heptane corresponds to $\frac{26}{100} = 0.26$ moles

$35$ g of octane corresponds to $\frac{35}{114} = 0.31$ moles

Mole fraction of heptane $χ = \frac{0.26}{0.26 + 0.31} = 0.456$

Mole fraction of octane $χ^{'} = 1 - 0.456 = 0.544$

Partial pressure of heptane $p = 0.456 \times 105.2 = 47.97$ kPa

Partial pressure of octane $p^{'} = 0.544 \times 46.8 = 25.46$ kPa

Vapour pressure of solution $P = p + p^{'} = 47.97 + 25.46 = 73.43$ kPa

Suggest Corrections

Similar questions

QUESTION 2.156

Heptane and octane form an ideal solution. At 373 K, the vapour pressure of the two liquid components is 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?

Q. Heptane and octane form ideal solution. At 373 K, the vapour pressures of the two liquids are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure, in bar, of a mixture of 25 g heptane and 35 g of octane?

Q. 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

k P a

and 45

k P a

respectively. Vapour pressure of the solution obtained by mixing 25.0 gm of heptane and 35 gm of octane will be:(molar mass of heptane

= 100 g m o 1^{- 1}

and of octane

= 114 g m o 1^{- 1}

)

Q. 1 mole of heptane (V.P.

= 92 m m

of Hg) was mixed with 4 moles of octane (V.P

= 31

mm of Hg). The vapour pressure of resulting ideal solution is:

Q. (a) The vapour pressure of n-hexane and n-heptane at

273 K

are

45.5 m m H g

and

11.4 m m H g

, respectively.What is the composition of a solution of these two liquids if its vapour pressure at

273 K

37.3 m m H g

.
(b) The mole fraction of n-hexane in the vapour above a solution of n-hexane and n-heptane is

0.75

273 K

.
What is the composition of the liquid solution?

Join BYJU'S Learning Program

Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?

Heptane and octane form an ideal solution. At $373 K$ , the vapour pressures of the two liquid components are $105.2 k P a$ and $46.8 k P a$ respectively. What will be the vapour pressure of a mixture of $26.0 g$ of heptane and $35 g$ of octane?