Multiple choice

The vapour pressure of water at $20$ is 17.5 mm Hg. If 18 g of glucose ${ C }{ 6 }{ H }{ 12 }{ O }_{ 6 }$ is added to 178.2 g water $20$ , the vapour pressure of the resulting solution will be:

  1. 17.675 mm Hg

  2. 15.750 mm Hg

  3. 16.500 mm Hg

  4. 17.325 mm Hg
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D Correct answer
Explanation

Relative lowering of vapour pressure = (P0 - Ps) / P0 = n_solute / (n_solute + n_solvent). Moles of glucose = 18/180 = 0.1. Moles of water = 178.2/18 = 9.9. (17.5 - Ps) / 17.5 = 0.1 / (0.1 + 9.9) = 0.1 / 10 = 0.01. 17.5 - Ps = 0.175. Ps = 17.325.

AI explanation

Using Raoult's law, the vapor pressure of the solution equals the mole fraction of the solvent multiplied by its pure vapor pressure. The moles of water are 178.2 g divided by 18 g/mol, giving 9.9 moles. The moles of glucose are 18 g divided by 180 g/mol, giving 0.1 moles, so the mole fraction of water is 9.9 divided by 10, which is 0.99. The final vapor pressure is 0.99 times 17.5 mm Hg, yielding 17.325 mm Hg.

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