CBSE (Science) Class 12CBSE
Share
Notifications

View all notifications

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. - CBSE (Science) Class 12 - Chemistry

Login
Create free account


      Forgot password?

Question

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?

Solution

Vapour pressure of heptane (`p_1^0`) = 105.2 kPa

Vapour pressure of heptane (`p_2^0`) = 46.8 kPa

We know that

Molar mass of heptane (C7H16)  = 7 x 12 + 16 x 1

= 100 g mol-1

∴Number of moles of heptane = `26/100` mol

= 0.26 mol

Molar mass of octane (C8H18) = 8 × 12 + 18 × 1

= 114 g mol−1

∴ Number of moles of octane = 35/114 mol

= 0.31 mol

Mole fraction of heptane, `x_1 = 0.26/(0.26 + 0.31)`

= 0.456

And, mole fraction of octane, x2 = 1 − 0.456

= 0.544

Now, partial pressure of heptane, `p_1=x_1p_1^0`

= 0.456 × 105.2

= 47.97 kPa

Partial pressure of octane `p_2 = x_2p_2^0`

= 0.544 × 46.8

= 25.46 kPa

Hence, vapour pressure of solution, ptotal p1 + p2

= 47.97 + 25.46

= 73.43 kPa

 

  Is there an error in this question or solution?

APPEARS IN

Video TutorialsVIEW ALL [1]

Solution 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. Concept: Ideal and Non-ideal Solutions.
S
View in app×