Question

How many grams of Kr are in a 9.30 L cylinder at 91.3 °C and 8.55 atm?

Answer 1

volume = 9.30 L

pressure = 8.55 atm

temperature = 91.3 C + 273 = 364.3 K

from ideal gas law

P V = n RT

8.55 x 9.30 = n x 0.0821 x 364.3

n = 2.66

moles of Kr = 2.66

molar mass of Kr = 83.798 g/mol

mass of Kr = moles x molar mass

                  = 2.66 x 83.798

                  = 222.78 g

mass of Kr = 222.78 g