Gaseous argon has a density of 1.40 g/L at standard conditions. How many argon atoms are in 1.00 L of argon gas at standard conditions? A)4.76 × 10²² B)3.43 × 10²⁵ C)2.11 × 10²² D)1.59 × 10²⁵ E)6.02 × 10²³

Question

Quizplus · Accepted Answer

To solve this problem, we need to use the ideal gas law, PV=nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the ideal gas constant, and T is temperature. At standard conditions (STP), pressure is 1 atm and temperature is 273 K. Rearranging the ideal gas law, we get:
n = PV/RT
We are given that the density of argon gas at standard conditions is 1.40 g/L. The molar mass of argon is 39.95 g/mol. Therefore, the number of moles of argon in 1 L of gas is:
n = (1.40 g/L)/(39.95 g/mol) = 0.035 L/mol
Plugging this into the ideal gas law, we get:
N = n * Avogadro's number = (0.035 L/mol) * (6.02 x 10^23/mol) = 2.11 x 10^22
Therefore, there are 2.11 x 10^22 argon atoms in 1.00 L of argon gas at standard conditions.

Gaseous Argon Has a Density of 1