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JEE Mains · Chemistry · STD 12 -1. Solution and colligative properties

\(8.5\; g \;CH _{2} Cl _{2}\) तथा \(11.95 \;g\; CHCl _{3}\) को मिलाकर एक विलयन तैयार किया जाता है। यदि \(298\; K\) पर \(CH _{2} Cl _{2}\) तथा \(CHCl _{3}\) के वाष्प दाब क्रमशः \(415\) तथा \(200 mmHg\) हो तो वाष्प रूप में उपस्थित \(CHCl _{3}\) का मोल अंश है। \(( Cl\) का मोलर द्रव्यमान \(=35.5 \;g \;mol ^{-1})\)

  1. A \(0.162\)
  2. B \(0.675\)
  3. C \(0.325\)
  4. D \(0.486\)
Verified Solution

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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