Calculating the molar volume of a gas using the ideal gas law formula PV = nRT:
P = 50.0 kPa = 50.0 x 10^3 Pa
V = unknown molar volume
n = 1 mole
R = 8.314 J/(mol∙K)
T = 300 K
Rearranging the formula to solve for V:
V = (nRT)/P
V = (1)(8.314)(300)/(50.0 x 10^3)
V = 24.8 L
Therefore, the molar volume of chlorine gas at 50.0 kPa and 300 K is 24.8 L.
The correct answer is 24.8 L.
The molar volume (i.e., volume that 1 mole of something takes up) of chlorine gas at 50.0 kPa and 300 K is
The molar volume (i.e., volume that 1 mole of something takes up) of chlorine gas at 50.0 kPa and 300 K is
49.9 L
45.1 L
24.8 L
70.9 L
22.4 L
1 answer