Apiston containing 0.120moles of methane gas, ch4, has a volume of 2.12liters. if methane is added until the volume is increased to 3.12liters with no change in pressure or temperature, how many grams are in the piston?

2.83 g

Explanation:

At constant temperature and pressure, Using Avogadro's law

Given ,  

V₁ = 2.12 L

V₂ = 3.12 L

n₁ = 0.120 moles

n₂ = ?

Using above equation as:

n₂ = 0.17660 moles

Molar mass of methane gas = 16.05 g/mol

So, Mass = Moles*Molar mass = 0.17660 * 16.05 g = 2.83 g

2.83 g  are in the piston.