What pressure will 2.6 x 1023 molecules of h2 exert in a 3.9 l container at 45â°c?

We can use the ideal gas law equation to find the pressure 
PV = nRT
where;
P - pressure
V - volume - 3.9 x 10⁻³ m³
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 45 °C + 273 = 318 K
n - number of moles 
there are 6.022 x 10²³ molecules in 1 mol of H₂
Therefore when theres 2.6 x 10²³ H₂ molecules - 1 / (6.022 x 10²³) x 2.6 x 10²³
Number of H₂ moles - 0.43 mol
substituting these values in the equation,
P x 3.9 x 10⁻³ m³ = 0.43 mol x  8.314 Jmol⁻¹K⁻¹ x 318 K
P = 291.5 kPa