Question:
Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.
Answer:
The vapour pressure of water, Po1 = 17.535 mm Hg
Given mass of glucose = 25 g
Given mass of water = 450 g
Calculate
Molar mass of glucose (C6H12O6) = 6 × 12 + 12 × 1 + 6 × 16 = 180 g /mol
Molar mass of water = 18 g/ mol
Number of moles of water = 450/18 = 25 mol
Number of moles of glucose = 25/180 = 0.139 mol
Using Raoult’s law Formula
(p10 - p1)/ p10 =(n2)/ (n1 + n2)
(17.535 - p1)/ 17.535=0.139/ 0.139 +25
(17.535 - p1) x 1.64 = 0.097 x 17.535
(17.535 - p1) = 0.097/1.64 = 0.06
p1 =17.535 - 0.06 = 17.47 mmHg
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