Question

the vapour pressure of pure water at 30 degree celsius is 31. 80 mm HG how many grams of Urea should be dissolved in hundred gram of water to lower the vapour pressure by 0.25 mm of HG​

Answer 1

Given:

The vapour pressure of pure water, P⁰ = 31. 80 mm Hg

Lowering of vapour pressure, P⁰- P = 0.25 mm Hg

Mass of water = 100 gm

To Find:

Amount of Urea required.

Calculation:

- We know that relative lowering of vapour pressure is equal to mole fraction of solute

⇒ (P⁰- P) / P⁰ = X

⇒ 0.25 / 31. 80 = (w/60)/ (w/60) + (100/18)

⇒ (w/60) + (100/18) / (w/60) = 31.80 / 0.25

⇒ 1 + (5.556 × 60) / w = 127.2

⇒ 333.36 / w = 126.2

⇒ w = 333.36 / 126.2

⇒ w = 2.64 gm

- So 2.64 gm grams of Urea should be dissolved in the given case.

Answer 2

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