Question

Answer the following question:

Answer 1

Answer: The mass of 45.0 L of $F_{2}$ at 87.0° C and 750 mm Hg is 56.605 g.

Explanation:

Given: Volume = 45.0 L

Temperature = $87.0^{o}C$ = (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.

$PV = nRT\\0.98 atm \times 45.0 L = n \times 0.0821 L atm/mol K \times 360 K\\n = \frac{0.98 atm \times 45.0 L}{0.0821 L atm/mol K \times 360 K}\\= \frac{44.1}{29.556} mol\\= 1.49 mol$

Moles is the mass of a substance divided by its molar mass. So, mass of $F_{2}$ (molar mass = 37.99 g/mol) is calculated as follows.

$Moles = \frac{mass}{molar mass}\\1.49 mol = \frac{mass}{37.99 g/mol}\\mass = 56.605 g$

Thus, we can conclude that the mass of 45.0 L of $F_{2}$ at 87.0° C and 750 mm Hg is 56.605 g.