Answer to Question #106062 in General Chemistry for Starlha Cañete

Question

How many grams of fluorine gas are in 3.28×10 to the fifth power L at a pressure of 5.79 kpa and a temperature of 46 degree celsius?
PV=nRT
Use R=0.80206 L atm/k mol

Answer 1

How many grams of fluorine gas are in 3.28×10⁵ L at a pressure of 5.79 kpa and a temperature of 46 ^o C?

PV=nRT

Use R=0.080206 L atm/k mol

Solution:

According to Mendeleev-Klapeyrone's equation:

pV = nRT (1)

or:

pV = mRT/M, where m is mass in grams; M - molar mass of substance in g/mol.

M(F₂) = 2*19 = 38 g/mol

Using equation (1) we can calculate the mass of F₂:

m(F₂) = pV*M/(RT)

where p is pressure which equals 5.79 kpa = 0.0571 atm;

T is temperature = 46^oC = 319 K

Thus:

m(F₂) = 0.0571*3.28*10⁵*38/(0.080206*319) = 27816.1 (g) = 27.816 (kg)

Answer: m(F₂) = 27816.1 grams.