Answer
0.515 atm
Work Step by Step
n= $\frac{1.55g}{131.3gmol^{-1}}$ = 0.012 mol
T= (20.+273) K = 293 K
V= 560 mL = .560 L
P = $\frac{nRT}{V}$ (using ideal gas law)
= $\frac{0.012mol(0.0821\frac{L.atm}{mol.K})293K}{.560 L}$ $\approx$ 0.515 atm
Chemistry 10th Edition
by Whitten, Kenneth W.; Davis, Raymond E.; Peck, Larry; Stanley, George G.
Published by
Brooks/Cole Publishing Co.
ISBN 10:
1133610668
ISBN 13:
978-1-13361-066-3
Chapter 12 - Gases and the Kinetic-Molecular Theory - Exercises - The Ideal Gas Equation - Page 442: 51
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