Aflashbulb of volume 2.70 ml contains o2(g) at a pressure of 2.30 atm and a temperature of 30.0 °c. how many grams of o2(g) does the flashbulb contain?

P = 2.30 atm

Volume in liter = 2.70 mL / 1000 => 0.0027 L

Temperature in K = 30.0 + 273 => 303 K

R = 0.082 atm

molar mass O2 = 31.9988 g/mol

number of moles O2 :

P * V = n * R* T

2.30 * 0.0027 = n * 0.082 * 303

0.00621 = n * 24.846

n = 0.00621 / 24.846

n = 0.0002499 moles of O2

Mass of O2:

n = m / mm

0.0002499 = m / 31.9988

m = 0.0002499 * 31.9988

m = 0.008 g

It is 4.92 e-4 sorry if i am wrong