JEE Mains · Chemistry · STD 12 -1. Solution and colligative properties
A solutivn is prepared by mixing \(8.5\,g\) of \(CH_2Cl_2\) and \(11.95\,g\) of \(CHCl_3\) . If vapour pressure of \(CH_2Cl_2\) and \(CHCl_3\) at \(298\,K\) are \(415\) and \(200\, mm\,Hg\) respectively, the mole fraction of \(CHCl_3\) in vapour form is : (Molar mass of \(Cl\, = 35.5\, g\, mol^{-1}\) )
- A \(0.162\)
- B \(0.675\)
- C \(0.325\)
- D \(0.486\)
Answer & Solution
Correct Answer
(C) \(0.325\)
Step-by-step Solution
Detailed explanation
Molar mass of \(CHCl_3\,=\,119.5\,g/mol.\) Molar mass of \(CH_2Cl_2\,=\,85\,g/mol.\) Molar of \(CHCl_3\,=\,\frac {11.95}{119.5}\,=\,0.1\,mol.\) Molar of \(CH_2Cl_2\,=\,\frac {8.5}{85}\,=\,0.1\,mol.\) Mole fraction of \(CHCl_3\,=\,\frac {0.1}{0.2}\,=\,0.5\,mol\) Mole fraction of…
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