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NTA Abhyas JEE Main2020ChemistrySolutionsPractice

18 glucose C 6 H 12 O 6 is added to 178.2 g water. The vapour pressure of water (in torr) for this aqueous solution at 273 K is

Options

  1. A759.0
  2. B739.6
  3. C746.0
  4. D752.4

Correct answer

D. 752.4

Step-by-step solution

Relative lowering of vapour pressure Equation P 0 - P s P 0 = X s o l u t e = n n + N Modified forms of equation is P 0 - P s P s = n N n = moles of solute, N = moles of solvent P 0 = 760 torr P s = ? 760 - P s P s = 18 180 178.2 18 P s = 752.4 torr

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