Answer: The number of gram of ammonia gas ([tex]NH_{3}[/tex]) contained in the a 3.0L vessel at 305 K with a pressure of 1.50 atm is 3.048 g.
Explanation:
Given: Volume = 3.0 L
Temperature = 305 K
Pressure = 1.50 atm
Formula used to calculate the number of 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\\1.50 atm \times 3.0 L = n \times 0.0821 L atm/mol K \times 305 K\\n = \frac{1.50 atm \times 3.0 L}{0.0821 L atm/mol K \times 305 K}\\= \frac{4.5}{25.0405}\\= 0.179 mol[/tex]
It is known that moles is the mass of substance divided by its molar mass.
Hence, mass of ammonia gas (molar mass = 17.03 g/mol) is as follows.
[tex]Moles = \frac{mass}{molar mass}\\0.179 mol = \frac{mass}{17.03 g/mol}\\mass = 3.048 g[/tex]
Thus, we can conclude that the number of gram of ammonia gas ([tex]NH_{3}[/tex]) contained in the a 3.0L vessel at 305 K with a pressure of 1.50 atm is 3.048 g.
