Determine the vapor pressure of a solution at 25degreeC that contains 76.6 g of

Question

Determine the vapor pressure of a solution at 25degreeC that contains 76.6 g of glucose (C6H1206) in 250.0 mL of water. The vapor pressure of pure water at 25degreeC is 23.8 torr. A) 70.8 torr B) 72.9 torr C) 23.1 torr D) 22.9 torr E) 7.29 torr 23) Determine the partial pressure of oxygen necessary to form an aqueous solution that is 7.1 x 10-4 M O2 at 25degreeC. The Henry's law constant for oxygen in water at 25degreeC is 1.3 x 10-3 M/atm. A) 1.1 atm B) 0.92 atm C) 0.88 atm D) 1.8 atm E) 0.55 atm Answer:

Explanation / Answer

22.

m = 76.6 g of sucrose

mol = mass/MW = 76.6/342.2965 = 0.22378 mol

V = 250 ml off water = 250 g

mol water = mass/MW = 250/18 = 13.8 mol

total = 13.8+0.22378 = 14.02378

x of sucrose = 0.22378 /14.02378 = 0.01595

then

dP = xsolute*P° = 0.01595*23.8 = 0.36685

Pfinal = 23.8 -0.36685 = 23.43315 mm HGnearest answer is C

2)

M = HP

7.1*10^-4 = (1.3*10^-3)(Pi)

Pi = (7.1*10^-4)/(1.3*10^-3) = 0.546 atm

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