Question

What is the vapor pressure of heptane in mm Hg at 98.4 oC over a solution that contains 0.500 mol of a non-volatile non-electrolyte dissolved in 200 grams of heptane?
[Useful Data: heptane (MW=100); normal boiling point = 98.4 oC ; Vapour pressure of pure heptane=760 mmHg]

Answer 1

Explanation:

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Answer 2

Given:

No of moles of non- volatile solute = 0.5

Mass of Heptane W = 200 gm

MW of heptane = 100 gm

Vapour pressure of pure heptane, P° = 760 mm Hg

To Find:

The vapor pressure of heptane in the solution at 98.4 °C.

Calculation:

- No of moles of heptane = 200 / 100 = 2

- Mole fraction of solute, X1 = 0.5 / (0.5 + 2)

⇒ X1 = 0.5 / 2.5

⇒ X1 = 0.2

- Mole fraction of Heptane, X2 = 1 - X1

⇒ X2 = 1 - 0.2

⇒ X2 = 0.8

- Vapour pressure of heptane is given by Raoult's law as follows:

P = P° X2

⇒ P = 760 × 0.8

⇒ P = 608 mm Hg

- So, the vapor pressure of heptane in the solution is 608 mm Hg.