What is the volume of 9.783 x 1023 atoms of kr at 9.25 atm and 512k?

1 mole of Kr contains Avogadro's number of atoms, therefore, to calculate the number of moles containing 9.783 * 10^23 atoms, we will do cross multiplication as follows:
number of moles = (9.783 * 10^23 *1) / (6.022 * 10^23) = 1.625 moles
1 atm is equivalent to 101325 Pa, therefore:
9.25 atm = 937256.25 Pa

The ideal gas law states that:
PV = nRT where
P is the pressure = 937256.25 Pa
V is the volume we want to calculate
n is the number of moles = 1.625 moles
R is the gas constant =  8.31441
T is the temperature = 512 k

Substitute with the givens to get V as follows:
937256.25 V = 1.625 * 8.31441 * 512
Volume = 7.38 * 10^-3 liters

Ithink its solid. hope this