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Unformatted text preview: f value of 1.86 C/m what would be the freezing point and boiling point of this solution? T f = K f * m T f = 1.86 * 0.14 * 2 T f = 0.52 Freezing point = -0.52 Celsius T b = K b * m T b = 0.512 * 0.14 * 2 T b = 0.14 Boiling point = 100.14 Celsius CHEM 1212 5. What is the osmotic pressure of the solution at 40.0 Celsius? = M * R * T = 0.15 M * 0.0821 L-atm/mol-K * T = 0.15 M * 2 * 0.0821 L-atm/mol-K * 313.15 = 7.71 atm 6. What is the vapor pressure of the solvent in this solution if the vapor pressure of pure water would be 92.5 torr at the temperature of the solution? P solvent = X solvent * P solvent X solvent = 73.36 mol / ( 2 * 0.1857 mol + 73.36 mol) = 0.9950 P solvent = 0.9950 * 92.5 torr P solvent = 92.0 torr Note, with vapor pressure, the number of ions factor, 2, goes with the moles of solute, which shows up in the denominator of the mole fraction....
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This note was uploaded on 11/29/2011 for the course CHEM 1212 taught by Professor Dockery during the Fall '08 term at Kennesaw.
- Fall '08