JEE Mains · Chemistry · STD 12 -1. Solution and colligative properties
A solution at \(20\,^oC\) is composed of \(1.5\,mol\) of benzene and \(3.5\,mol\) of toluene. If the vapour pressure of pure benzene and pure toluene at this temperature are \(74.7\,torr\) and \(22 .3\, torr,\) respectively, then the total vapour pressure of the solution and the benzene mole fraction in equilibrium with it will be, respectively
- A \(35.8\,torr\) and \(0.280\)
- B \(38.0\,torr\) and \(0.589\)
- C \(30.5\,torr\) and \(0.389\)
- D \(30.5\,torr\) and \(0.480\)
Answer & Solution
Correct Answer
(B) \(38.0\,torr\) and \(0.589\)
Step-by-step Solution
Detailed explanation
Total \(V.P.\) of solution \(=\,P^o_Ax_A\,+\,P^o_Bx_B\) Given, \(P^o_A\,=\,74.7\,torr\), \(P^o_B\,=\,22.3\,torr\) \(n_{benzene}\,=\,1.5\,mol\), \(n_{toluene}\,=\,3.5\,mol\) \(n_{solution}\,=\,1.5+3.5=5\,mol\) \(X_A\,=\,\frac {n_{benzene}}{n_{solution}}\,=\,\frac {1.5}{5}\) Total…
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