JEE Mains · Chemistry · STD 12 -1. Solution and colligative properties
A non-volatile, non-electrolyte solid solute when dissolved in \(40\) g of a solvent, the vapour pressure of the solvent decreased from \(760\) mm Hg to \(750\) mm Hg. If the same solution boils at \(320\) K, then the number of moles of the solvent present in the solution is _____. (Nearest integer)
[Given: boiling point of the pure solvent \(= 319.5\) K, \(K_b\) of the solvent \(= 0.3\) K kg mol\(^{-1}\)]
- A 5
- B 10
- C 15
- D 20
Answer & Solution
Correct Answer
(A) 5
Step-by-step Solution
Detailed explanation
From the relative lowering of vapour pressure, we have: \(\dfrac{P^0 - P_s}{P_s} = \dfrac{n_B}{n_A}\) where \(n_B\) is the number of moles of solute and \(n_A\) is the number of moles of solvent. Substituting the given values: \(\dfrac{760 - 750}{750} = \dfrac{n_B}{n_A}\)…
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