Question

Which of the following is the mathematical form of Raoult's law?

• A. \( \dfrac{P^{0}-P}{P^{0}} = \dfrac{n}{n+N} \)
• B. \( \dfrac{P^{0}-P}{P^{0}} = \dfrac{N}{n+N} \)
• C. \( \dfrac{P^{0}-P}{P} = \dfrac{n}{N+n} \)
• D. \( \dfrac{P-P^{0}}{P^{0}} = \dfrac{n}{n+N} \)

Hint:

Relative lowering of vapour pressure equals the mole fraction of the solute, \( n/(n+N) \).

Answer 1

The Correct Option is A

Step 1: For a solution of a non-volatile solute, Raoult's law states that the relative lowering of vapour pressure equals the mole fraction of the solute.
Step 2: If \( P^{0} \) is the vapour pressure of pure solvent and \( P \) that of the solution, then \( \dfrac{P^{0}-P}{P^{0}} = x_{solute} = \dfrac{n}{n+N} \), where \( n \) = moles of solute and \( N \) = moles of solvent.
Step 3: This is exactly option (i), so option 1 is correct.
Step 4: Why the others are wrong: option (ii) uses \( N \) (solvent) in the numerator, which is the mole fraction of solvent; option (iii) divides by \( P \) instead of \( P^{0} \); option (iv) reverses the sign giving a negative value.
Answer: \( \dfrac{P^{0}-P}{P^{0}} = \dfrac{n}{n+N} \).