Answer : The mass of ammonia present in the flask in three significant figures are, 5.28 grams.
Solution :
Using ideal gas equation,
[tex]PV=nRT\\\\PV=\frac{w}{M}\times RT[/tex]
where,
n = number of moles of gas
w = mass of ammonia gas = ?
P = pressure of the ammonia gas = 2.55 atm
T = temperature of the ammonia gas = [tex]27^oC=273+27=300K[/tex]
M = molar mass of ammonia gas = 17 g/mole
R = gas constant = 0.0821 L.atm/mole.K
V = volume of ammonia gas = 3.00 L
Now put all the given values in the above equation, we get the mass of ammonia gas.
[tex](2.55atm)\times (3.00L)=\frac{w}{17g/mole}\times (0.0821L.atm/mole.K)\times (300K)[/tex]
[tex]w=5.28g[/tex]
Therefore, the mass of ammonia present in the flask in three significant figures are, 5.28 grams.
