# Molar Mass of Unknown Solute and Mole Fraction of Gases

by Soaring Crane
Tags: fraction, gases, mass, molar, mole, solute, unknown
 P: 483 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.
 Admin P: 21,583 I think the approach to a) is correct, but let me get GCT or Ouabache to take a look.
P: 166

## Molar Mass of Unknown Solute and Mole Fraction of Gases

#1 is correct
 P: 483 Is #2 correct? Thanks again.
 P: 17 #2 is correct too.
 Sci Advisor HW Helper P: 1,769 yeah, both of them seem fine.
 Sci Advisor HW Helper P: 1,327 Nice job Soaring Crane.. I also tried them and come up with those solutions.

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