Question

A sample of wet argon is collected in a 2.46 L container at 775 torr and 30.2 °C. It was subsequently found that there were 0.0480 moles of water vapor in this sample. How many moles of argon gas are in the sample? Report your answer to the correct number of significant figures.

Answer 1

Answer

0.053 moles of Ar gas in the sample

Explanation

To find the number of moles of wet Ar
T = 303.2 K

P = 1.0197 atm

R = 0.08205 L atm mol-1 K-1

V = 2.46 L

PV = nRT

n = PV/RT

   = $\frac{(1.0197atm)*(2.46L)}{(0.08205Latmmol^{-1}K^{-1})*(303.2K)}$

   = 2.5085/24.878

   = 0.1008 moles

Since the collected gas is wet, it will have water molecules mixed with it.

number of moles of wet Ar = number of moles of dry Ar + number of moles of water

number of moles of dry Ar = number of moles of wet Ar - number of moles of water

number of moles of dry Ar = 0.1008 - 0.0480

                                      = 0.0528 moles (or) 0.053 moles