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Question 16

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?

Answer

Vapour pressure of heptane p10 = 105.2 kPa

Vapour pressure of octane p20= 46.8 kPa

As we know that, Molar mass of heptane (C7H16) = 7 × 12 + 16 × 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, x1 = 0.26 / 0.26 +0.31

= 0.456

And, mole fraction of octane, x2 = 1 - 0.456 = 0.544

Now, partial pressure of heptane, p1 = x2 p20

= 0.456 × 105.2

= 47.97 kPa

Partial pressure of octane,p2 = x2 p20

= 0.544 × 46.8 = 25.46 kPa

Hence, vapour pressure of solution, ptotal  =  p1 + p2

= 47.97 + 25.46

= 73.43 kPa

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