ExamBro
ExamBro
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

  1. A \(\frac{3}{4}\)
  2. B \(\frac{1}{2}\)
  3. C \(\frac{1}{4}\)
  4. D \(\frac{4}{3}\)
Verified Solution

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…
Same subject
Explore more questions on app