p = 1.67*10^4 kilo Pascals
V = 43.8 L
T = 273 + 24 = 297 K
R - 8.314 Liters kilopascals/deg K
PV = n R T
n = (1.67*10^4)(43.8)/ [ 8.314*297 ]
n = 296 mols
296 * 40 grams/mol = 11840 grams
or 11.84 kg
Given that many laboratory gases are sold in steel cylinders with a volume of 43.8 L, What mass (in grams) of argon is inside a cylinder whose pressure is 17810kPa at 24∘C?
I've attempted this problem but I'm not getting the correct answer. Please explain if you can!
1 answer