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Solutions question

2018 · 15 Apr · Shift 2 · Q12
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Solutions question

2018 · 15 Apr · Shift 2 · Q12

JEE MainChemistrySolutionsMCQ+4 / −1
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 Mx and My, respectively where Mx = 34{3 \over 4}43​ My. The relative lowering of vapor 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 the solvent, the value of ''m'' is :
  1. A
    43{4 \over 3}34​
  2. B
    34{3 \over 4}43​
  3. C
    12{1 \over 2}21​
  4. D
    14{1 \over 4}41​
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