Solutions

Solutions

Question

18 g glucose (C6H12O6 ) is added to 178.2 g water. The vapor pressure of water (in torr) for this aqueous solution is:

  • 76.0

  • 752.4

  • 759.0

  • 7.6

Answer

B.

752.4

Vapour pressure of water (p°) = 760 torr
Number of moles of glucose
equals space fraction numerator Mas space left parenthesis straight g right parenthesis over denominator Molecular space mass space left parenthesis straight g space mol to the power of negative 1 end exponent right parenthesis end fraction

equals space fraction numerator 18 space straight g over denominator 180 space straight g space mol to the power of negative 1 end exponent end fraction space equals space 0.1
Molar mass of water = 18 g/mol
Mass of water (given) = 178.2 g
Number of moles of water

equals space fraction numerator Mass space of space water over denominator Molar space mass space of space water end fraction space equals space fraction numerator 178.2 over denominator 18 space straight g end fraction space equals space 9.9 space mol
Total number of moles = (0.1 +9.9) = 10 moles
Now, the mole fraction of glucose in solution = Change in pressure with respect to initial pressure.
straight i. straight e. space fraction numerator increment straight p over denominator increment straight p to the power of straight o end fraction space equals space fraction numerator 0.1 over denominator 10 end fraction
or space increment straight p space equals space 0.01 space straight p to the power of straight o space equals space 0.01 space straight x space 760 space equals space 7.6 space torr
therefore, Vapour pressure of solution = (760-7.6)torr
=752.4 torr

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