Question

The density of chlorine gas at 0.970 atm and 29.8 °C is ________ g/L.

Answer 1

The density of chlorine gas at the given conditions can be calculated using the Ideal Gas Law formula by rearranging it to express density as a function of pressure, molar mass, the gas constant, and temperature.

The density of chlorine gas at 0.970 atm and 29.8 °C can be calculated using the Ideal Gas Law PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin. The density (ρ) is mass (m) divided by volume (V), or ρ = m/V. Since the number of moles (n) can be expressed as mass (m) divided by the molar mass (M), we can rewrite the equation as PV = (m/M)RT. Rearranging for m/V gives us ρ = (PM)/(RT). Substituting the given values into this equation—P = 0.970 atm, M = 70.906 g/mol for chlorine, R = 0.0821 L·atm/K·mol, and T = 273.15 + 29.8 K—we can calculate the density.

to find the density of chlorine gas, one must use the provided atmospheric conditions in conjunction with the known molar mass of chlorine and the Ideal Gas Law to arrive at the solution.

Answer 2

d=3.95 g/l

Step-by-step explanation:

1) Ideal gas equation:

pV = nRT

2) Transform the equation to compute density, d = m / V

n = mass in grams / molar mass = m / MM

pV = (m/MM) RT

=> pV = mRT/MM

=> m/V = pMM / (RT)

d = pMM / (RT)

p = 1.21 atm

MM = 83.80 g/mol (this is the atomic mass of krypton element)

T = 50 + 273.15 K = 323.15K

3) Compute:

d = (1.25 atm * 83.80 g/mol) / (0.0821 atm*liter /K*mol * 323.15K)