A 0.50 L container at 305 K holds a 0.22 g sample
of carbon dioxide gas (CO2). Assuming ideal
gas behavior, what is the pressure, in kilopascals,
exerted by the gas?
A 23 kPa
B 25 kPa
C 41 kPa
D 67 kPa

Option B. 25 kPa

Explanation:

First, let us calculate the number of mole of CO2 in the container. This is illustrated below:

Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol

Mass of CO2 = 0.22g

Number of mole of CO2 = 0.22/44 = 0.005mol

From the question, we obtained the following data:

V = 0.5L

T = 305K

R = 0.082atm.L/K /mol

n = 0.005mol

P =?

PV = nRT

P = nRT/V

P = (0.005x0.082x305)/0.5

P = 0.2501atm

Recall:

1atm = 101325Pa

0.2501atm = 0.2501 x 101325 = 25341.4Pa = 25341.4/ 1000 = 25KPa

narendra modi is currently the pm of india

explanation: