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9 gram anhydrous oxalic acid (mol. Wt. = 90) was dissolved in 9.9 moles of water. If vapour pressure of pure water is P 1 o c the vapour pressure of solution is

Options

  1. A0.99 P 1 o
  2. B0.1 P 1 o
  3. C0.9 1 P 1 o
  4. D1.1 P 1 o

Correct answer

A. 0.99 P 1 o

Step-by-step solution

The total vapour pressure of a solution in this case only depends on vapour pressure of water as anhydrous oxalic acid is a non-volatile compound. ∴ Vapour pressure of solution = vapour pressure of Water P w According to Raoult’s law Number of moles of oxalic acid = 9 90 = 0.1 moles ∴ x w = 9.9 9.9 + 0.1 = 0.99 ⇒ P s = P w = 0.99 × P 1 o

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