JEE Mains · Chemistry · STD 12 -1. Solution and colligative properties
When \(0.25\) moles of a non-volatile, non-ionizable solute was dissolved in \(1\) mole of a solvent the vapor pressure of solution was \(x\%\) of vapor pressure of pure solvent. What is \(x\%\)?
- A \(50\%\)
- B \(60\%\)
- C \(70\%\)
- D \(80\%\)
Answer & Solution
Correct Answer
(D) \(80\%\)
Step-by-step Solution
Detailed explanation
Moles of solute, \(n_B = 0.25\) Moles of solvent, \(n_A = 1\) Mole fraction of solvent, \(X_A = \dfrac{n_A}{n_A + n_B} = \dfrac{1}{1 + 0.25} = \dfrac{1}{1.25} = 0.8\) According to Raoult's law, the vapor pressure of the solution is given by \(P_s = P^0 X_A\)…
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