How many grams of argon would it take to fill a light bulb with a volume of 0.475 l at stp?

STP stands for Standard Temp. and Pressure. 
Standard Temp. = 273.15 K
Pressure = 1 atm
Now,  ideal gas equation we know that PV = nRT
where, V = volume of gas = 0.475 l
R = universal gas constant = .082507 l atm mol-1 K-1 

∴ n =
∴ n =
∴n = 0.02112 mol

Thus, number of moles of Ar present = 0.02112

But, number of mole = 
In present case, weight of Ar (g) = number of mole of Ar X Atomic Wt. of Ar
∴ weight of Ar = 40 X 0.02112 = 0.8477 g

The answer to this problem is. 67