Question

2.98 g of CO

gas has a volume of 4190 mL at 19.2 Celsius what is atm

Answer 1

Given:

$m_{CO}=2.98 g$

$V=4.19l$

$T=19.2 C = 292.2K$

$R=0.082 ( \tfrac{L*atm}{K*mol})$

Solution:

n(mole of CO) = $\tfrac{2.98}{28} =0.106mole$

Ideal gas law:

$PV= nRT$

$P= \tfrac{ nRT}{V}$

$P= \tfrac{0.106*0.082*292.2}{4.19} =0.608$ atm

P = 0.608 atm