Question

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

पर्याय

• 7.6

• 76.0

• 752.4

• 759.0

Answer 1

18 g glucose (C₆H₁₂O₆) is added to 178.2 g water. The vapour pressure of water (in torr) for this aqueous solution is 752.4.

Explanation:

Moles of glucose = `18/180` = 0.1

Moles of water = `178.2/18` = 9.9

Total moles = 0.1 + 9.9 = 10

`"p"_("H"_2"O")` = Mole fraction × total pressure

= `9.9/10 xx 760`

= 752.40 Torr