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Unformatted text preview: 3 P)/(10 6 cm 3 air) = (x cm 3 P)/(25 cm 3 air) Solving for x gives 3.0 x 104 cm 3 Example5: If P has a molecular weight of 36 g mol1 , how many grams of P is present in this sample? Assume 1 atm and 25 degC. We know the volume of P from Example3. If we can determine how many molecules of P occupy this volume, then using the molecular weight, we can determine the mass of P. To convert the volume of P to moles of P, use the ideal gas law: PV = nRT, where P = 1 atm, T = 25 + 273 = 298 deg. K, and R = 0.0821 L atm mol1 K1 . n = PV/RT = (1 atm) (3.0 x 107 L) /{(0.082 L atm mol1 K1 )(298 K)} = 1.2 x 108 mol P When using the Ideal Gas Law, dont forget to express the volume in L. Recall that 1L = 1000 cm 3 , so 3.0 x 104 cm 3 is equivalent to 3.0 x 107 L. From molecular weight of P, compute final answer: (1.2 x 108 mol P)(36 g mol P1 ) = 4.3 x 107 g T. Komada 1/26/2010...
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 Spring '11
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 Chemistry, Partial Pressure, pH

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