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Topic: Molar mass of an unknown substance if 227.5 mL of an aqueous solution  (Read 3702 times)

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Offline Sis290025

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1) Find the molar mass of an unknown substance if 227.5 mL of an aqueous solution containing 2.785 grams of the unknown generates an osmotic pressure of 588.0 mm Hg at 20.9 Celsius. Report your answer to 3 significant figures.

pi = MRT

pi = 588.0 mm Hg = 588 mm Hg/760 mm Hg = 0.7736842 atm
T = 20.9 C = 20.9 C + 273.15 = 294.05 K

molar mass = g solute / mol solute

pi = R*T*(mol solute/L solution)

mol solute = (pi*Volume_solution)/(R*T) = (0.7736842 atm * 0.2275 L)/(294.05 K * 0.08206) = 0.007294448 mol

m. mass = 2.785 g / 0.007294448 mol = 381.797 g/mol = 382 g/mol ??


2) Calculate the mole fraction of benzene in the vapor phase from a solution that contains a mole fraction of benzene of 0.557 when mixed with toluene if the vapor pressure of pure benzene is 183 mmHg and that of toluene is 59.2 mm Hg.

X_b = 0.557-------------------VP_b = 183 mm Hg
X_t = 1 - 0.557 = 0.443-------VP_t = 59. 2 mm Hg

Total (vapor) pressure = (X_b*P_b) + (X_t*P_t)
P_t = 0.557*183 mm Hg + 0.443*59.2 mm Hg
P_t = 101.931 mm Hg + 26.2256 mm Hg = 128.1566 mm Hg

X_gas = P_gas (benzene) / P_total = 101.931 mm Hg / 128.1566 mm Hg = 0.795 ?


Thank you.
« Last Edit: January 29, 2006, 07:14:17 PM by Mitch »

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