what is the pressure of 9000000000000000000 molecules/ml of NO at 273 K in 1L?

Answer 1

(0.09/6.022)*(22.414/6.022) atm

0.09 /6.022= n= no. of moles (n=no. of molecules/avogadro no.)
avogadro no. =6.022*10^23

Answer 2

If I'm counting right, that's 9 x 10^18 molecules/ml, so in 1 L that would be 9 x 10^21 molecules, or 0.015 moles.

Plug everything else into the Ideal Gas Law and solve for pressure.

P = nRT/V = (0.015)(0.08206)(273)/(1) = 0.336 atm