What is the density of 0.50 grams of gaseous carbon stored under 1.5 atm of pressure at a temperature of - 20.0 C?

Answer: The density of 0.50 grams of gaseous carbon stored under 1.50 atm of pressure at a temperature of - 20.0 °C is 0.867 g/L.

Explanation:

d = m/V, where d is the density, m is the mass and V is the volume. We have the mass m = 0.50 g, so we must get the volume V. To get the volume of a gas, we apply the general gas law PV = nRT

P is the pressure in atm (P = 1.5 atm)

V is the volume in L (V = ? L)

n is the number of moles in mole, n = m/Atomic mass, n = 0.50/12.0 = 0.416 mole.

R is the general gas constant (R = 0.082 L. atm/mol. K).

T is the temperature in K (T (K) = T (°C) + 273 = - 20.0 + 273 = 253 K).

Then, V = nRT/P = (0.416 mol) (0.082 L. atm/mol. K) (253 K) / (1.5 atm) = 0.576 L. Now, we can obtain the density; d = m/V = (0.50 g) / (0.576 L) = 0.867 g/L.