3.6866 × 1025 particles of PH3 is held at 675.02 kPa and 524 °C. What is the volume of its container in milliliters?

To solve this problem, we can use the ideal gas law equation, which is PV = nRT.

First, we need to convert the given conditions to the appropriate units.

1. Pressure (P) = 675.02 kPa = 675.02 x 10^3 Pa (since 1 kPa = 10^3 Pa)
2. Temperature (T) = 524 °C = 524 + 273.15 = 797.15 K (since temperature in the ideal gas law equation should be in Kelvin)
3. The number of moles (n) can be calculated using Avogadro's number (6.022 x 10^23 particles/mole). So, 3.6866 x 10^25 particles of PH3 = 3.6866 x 10^25 / 6.022 x 10^23 = 61.24 moles.
4. The ideal gas constant (R) = 8.314 J/(mol.K)

Now, we can substitute these values into the ideal gas law equation to find the volume (V).

V = nRT/P = (61.24 moles * 8.314 J/(mol.K) * 797.15 K) / (675.02 x 10^3 Pa) = 0.596 m^3

Since 1 m^3 = 10^6 mL, the volume of the container is 0.596 m^3 * 10^6 = 596,000 mL.