Question

multiple choice Q

How many molecules of N₂ are in a 400.0 mL container at 1.0399 bar and 135 °C?

7.38 × 10²¹ molecules

7.01 × 10²¹ molecules

2.12 × 10²² molecules

2.23 × 10²² molecules

Answer 1

We know that ideal gas equation is PV = nRT

Where

T = Temperature = 135 oC =135+273 = 408 K

P = pressure = 1.0399 bar = 1.0399x0.987 atm since 1 bar = 0.987 atm

                    = 1.03 atm

n = No . of moles = ?

R = gas constant = 0.0821 L atm / mol - K

V= Volume of the gas = 400.0 mL = 0.4 L

Plug the values we get n = ( PV) / (RT)

                                    = 0.0122 moles

We know that 1 mole of any gas contains avagadro number = 6.023x10²³ molecules

0.0122 moles of N₂ gas contains M molecules

M = ( 0.0122x6.023x1023) / 1

    = 7.38x10²¹ molecules