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JEE Mains · Chemistry · STD 12 -1. Solution and colligative properties
Two \(5\) molal solutions are prepared by dissolving a non-electrolyte, non-volatile solute separately in the solvents \(X\) and \(Y\). The molecular weights of the solvents are \(M_X\) and \(M_Y\), respectively where \(M_X\, = \frac{3}{4} M_Y\). The relative lowering of vapour pressure of the solution in \(X\) is \(''m''\) times that of the solution in \(Y\). Given that the number of moles of solute is very small in comparison to that of solvent, the value of \(''m''\) is
- A \(\frac{3}{4}\)
- B \(\frac{1}{2}\)
- C \(\frac{1}{4}\)
- D \(\frac{4}{3}\)
Answer & Solution
Correct Answer
(A) \(\frac{3}{4}\)
Step-by-step Solution
Detailed explanation
The relationship between molar masses of the two solvents is \(M_X\,=\,\frac {3}{4}M_Y\) .... \((i)\) The relative lowering of vapour pressure of two solution is \({\left( {\frac{{\Delta P}}{P}} \right)_X}\, = \,m{\left( {\frac{{\Delta P}}{P}} \right)_Y}\) But, the relative…
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