150 g NaCL completely dissolved (Producing Na^+ and Cl^- ions) in 1.00 kg of wat

Question

150 g NaCL completely dissolved (Producing Na^+ and Cl^- ions) in 1.00 kg of water at 25degreeC. The vapor pressure of pure water at this temperature is 23.8 torr. What is the vapor pressure of the solution? The vapor pressure above an aqueous sugar solution at 25 degreeC is 21.9 torr. What is the mole fraction of sugar in this solution? (Assume sugar is a nondissociating, nonvolatile solute and that the vapor pressure of water 25 degreeC is 23.8 torr.) Gas X has a solubility of 0.25 g/L under a pressure of 9.8 atm. What solubility will it have under a pressure of 5.3 atm?

Explanation / Answer

16. raoults law

P0 - Ps/P0 = i*Xb

(23.8-x)/x = 2*((150/58.5)/((150/58.5)+(1000/18))

x = vaor pressure of solution = 21.87 torr

17.

P0 - Ps/P0 = i*Xb

(23.8-21.9)/21.9 = 1*x

x = molfraction of Sugar = 0.086

18.

C1p1 = c2p2

0.25*9.8 = c*5.3

c= solubility at 5.3 atm = 0.46 g/ml

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