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Heat Of Formation & Gas Laws

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An ideal gas at 300K and 2.00 atm pressure in a 400 cm^3 container has its pressure increased to 5.00 atm and its volume increased to 6000 cm^3. What is the final temperature?

The vapor pressure of water at 23 degrees celcius is 21.07 torr. Find the vapor pressure at 23 degrees celcius of the solution formed by dissolving 15.0 g o C6H12O6 (a nonvolatile solid) in 72.0 g of H2O.

The specific heat of ice is 2.09 Jg-K and the specific heat of liquid water is 4.18 J/g-K. The heat of fusion of ice is 6.01 kj/mol. Find the heat needed to convert 36.0 g of ice at -10 degrees celcius to liquid water at 20 degrees celcius.

The freezing point of C6H6 is 5.50 degrees celcius and Kf = 5.12 degrees celcius (mol/kg) for C6H6. A solution of 6.42 g of the nonelectrolyte compound X in 183.2 g of C6H6 freezes at 4.00 degrees celcius Find the molecular weight of X.

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Solution Summary

Four problems based on gas laws, vapor pressure, heat of formation and molality are solved.

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1. P1V1 / T1 = P2V2 /T2

T2 = P2V2T1 / P1V1 = 5 x 6000 x 300 / (2 x 400) = 11250 K

2. Vapor pressure of solution = mole fraction of solvent x pure vapor pressure of solvent

Moles of solvent = 72 / 18 = 4 ...

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