Question

A 22.6 gram sample of krypton gas has a volume of 904 milliliters at a pressure of 2.71 atm. The temperature of the Kr gas sample is:_____

Answer 1

Answer:

110.78 K

Explanation:

First we convert 22.6 g of krypton gas (Kr) to moles, using its molar mass:

• 22.6 g Kr ÷ 83.798 g/mol = 0.270 mol Kr

Then we use the PV=nRT formula, where:

• P = 2.71 atm

• V = 904 mL ⇒ 904/1000 = 0.904 L

• n = 0.270 mol

• R = 0.082 atm·L·mol⁻¹·K⁻¹

• T = ?

2.71 atm * 0.904 L = 0.270 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * T

And solve for T:

• T = 110.78 K