JEE Mains · Chemistry · STD 12 -1. Solution and colligative properties
A substance ' X ' \((1.5 g)\) dissolved in 150 g of a solvent ' Y ' (molar mass \(=300 g mol ^{-1}\) ) led to an elevation of the boiling point by 0.5 K . The relative lowering in the vapour pressure of the solvent ' Y ' is __________ \(\times 10^{-2}\). (Nearest integer) [Given : \(K _{ b }\) of the solvent \(=5.0 K kg mol ^{-1}\) ] Assume the solution to be dilute and no association or dissociation of X takes place in solution.
- A 3
- B 1
- C 5
- D 2
Answer & Solution
Correct Answer
(A) 3
Step-by-step Solution
Detailed explanation
\(\Delta T _{ b }= i \times K _{ b } \times m\) \(0.5= i \times m \times 5\) \(i \times m =\frac{0.5}{5}=0.1\) \(i \times a =\frac{15}{1000}\) (where \(a=\) moles of solute) Now, \(\frac{P_o-P_S}{P^o}=i X_{\text {solute }}=i \times \frac{a}{a+\frac{150}{300}}\)…
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