Answer: The mass of 45.0 L of [tex]F_{2}[/tex] at 87.0° C and 750 mm Hg is 56.605 g.
Explanation:
Given: Volume = 45.0 L
Temperature = [tex]87.0^{o}C[/tex] = (87.0 + 273) K = 360 K
Pressure = 750 mm Hg (1 mm Hg = 0.00131579 atm) = 0.98 atm
Formula used to calculate moles is as follows.
PV = nRT
where,
P = pressure
V = volume
n = no. of moles
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the values into above formula as follows.
[tex]PV = nRT\\0.98 atm \times 45.0 L = n \times 0.0821 L atm/mol K \times 360 K\\n = \frac{0.98 atm \times 45.0 L}{0.0821 L atm/mol K \times 360 K}\\= \frac{44.1}{29.556} mol\\= 1.49 mol[/tex]
Moles is the mass of a substance divided by its molar mass. So, mass of [tex]F_{2}[/tex] (molar mass = 37.99 g/mol) is calculated as follows.
[tex]Moles = \frac{mass}{molar mass}\\1.49 mol = \frac{mass}{37.99 g/mol}\\mass = 56.605 g[/tex]
Thus, we can conclude that the mass of 45.0 L of [tex]F_{2}[/tex] at 87.0° C and 750 mm Hg is 56.605 g.
