JEE Mains · Chemistry · STD 12 -1. Solution and colligative properties
The vapour pressure of acetone at \(20\,^oC\) is \(185\,torr.\) When \(1.2\,g\) of a non-volatile substance was dissolved in \(100\,g\) of acetone at \(20\,^oC,\) its vapour pressure was \(183\,torr.\) The molar mass \((g \,mol^{-1})\) of the substance is :
- A \(128\)
- B \(488\)
- C \(32\)
- D \(64\)
Answer & Solution
Correct Answer
(D) \(64\)
Step-by-step Solution
Detailed explanation
\(\frac{P^{o}-P_{x}}{P_{x}}=\frac{w_{2} M_{1}}{w_{1} M_{2}}\) where \(w_{1}, \,M_{1}=\) mass in \(g\) and mol. mass of solvent \(w_{2}, M_{2}=\) mass in \(g\) and mol. mass of solute Let \(M_{2}=x\) \(P^{o}=185\,torr\) ; \(P_{s}=183\,torr\)…
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