8) At a given temperature, the vapor pressures of hexane and octane are 183 mm H

Question

8) At a given temperature, the vapor pressures of hexane and octane are 183 mm Hg and 59.2 mm Hg, respectively. Calculate the total vapor pressure over a solution of hexane and octane. The mole fraction of hexane (Xhexane) is 0.580. 31 ?. 102.0 A) 106 mm Hg 131 mm Hg 121 mm Hg D) 242 mm Hg 103 34.33 9) A 200.0 mL sample of an aqueous solution at 25°C contains 30.2 mg of an unknown nonelectrolyte compound. If the solution has an osmotic pressure of 10.44 torr, what is the molar mass of the unknown compound? A) 294 g/mol 69 g/mo C) 341 g/mol D) 448 g/mol E) 195 g/mol 2 1 10) What is the freezing point of a solution of 7.15 g MgCl2 in 100 g of water? Kffor water is 1.86°C/m A)-4.18°C -0.140°C 40°C D) -2.80°C 11) Determine the solubility of N2 in water exposed to air at 25 C if the atmospheric pressure is 2.0 atm. Assume that the mole fraction of nitrogen is 0.78 in air and the Henry's law constant for nitrogen in water at this temperature is 6.1 × 10-4 Matm. A)1.6×10-4 M B) 6.4 x 10-4 M C) 1.2 x 104 M D)1.1×10-4 M P-(75)(2) 9.5 x 104M 12) Which one of the following will form an acidic solution in water? A) LiF B) Nal ONH4CI D) L?NO3 E) None of the above solutions will be acidic.

Explanation / Answer

1. We know that according to the Raoults law PA= xA * P0

Given X hexane = 0.580
Thus X octane = 1-0.580 = 0.420
vapor pressure of hexane = 183 mmHg
Vapor pressure of octane = 59.2 mmHg
Thus Phexane = Xhexane * P
Thus Phexane = 0.580* 183 = 106.14 mmHg
P octane = 0.420* 59.2 = 24.864 mmHg
Total pressure = Phexane + P octane = 106.14 + 24.864 = 131 mmHg

2. We find the Molarity using osmotic pressure and temperature
p=MRT
p = 10.44 torr = 10.44/760 = 0.013737 atm
M = P / RT = 0.013737 / (0.0821 L atm/mol K) *(298K)) = 0.0005614 M
Now Molarity = Moles / liters of solution
Moles = 0.0005614 * 0.2 = 1.12 * 10-4 moles
Thus molar mass = grams / Moles = 30.2 * 10-3 / 1.12 * 10-4 = 269 g/mol

3. We know that Change in T = i x m x kf
Where i is the number of particles the compound dissociates, m is the molality, and kf is the constant specific to water.
Now Molecular Mass of MgCl2 = 95.21 g/mol
and MgCl2 form three atoms ie i = 3 because 1Mg + 2Cl
moles of MgCl2 = 7.15 g /95.21 = 0.0751
Molality = moles/ mass of solvent in kG = 0.0751/0.1 = 0.751
Change in T = (3)*(0.751)*(1.86) = 4.18 C
Now we know that freezing point of water = 0 C
Thus freezing point of solution = 0-4.18 = -4.18 C

4. Solubility= K(Henrys Constant) x Partial Pressure of Nitrogen
Partial Pressure= (Mole Fraction of Nitrogen) x (Total Atmospheric Pressure)
Partial Pressure of Nitrogen = (.78 moles of Nitrogen) x (2 atm) = 1.56
Thus Solubility of Nitrogen= (6.1x10^-4) x (1.56) = 9.516x10^-4

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