What is the mole percent of neon on the mixture?

Question

A glass vessel fitted with a stopcock has a mas of 337.428g when exacuated. When fillied with argon, it has a mass of 339.854g. When evacuated and refilled with a mixture of argon and neon, under the same conditions of temperature and pressure, it weighs 339.076g. What is the mole percent of neon on the mixture?

please explain.

Answer 1

In the first run you filled with
m = 339.854g - 337.428g = 2.426g of pure Argon.which represent
N = m / M(Ar) = 2.426g / 39.948g/mol = 0.06073 mol

Assume the gas in the vessel behaves like an ideal gas. Then the amount in the vessel is given by:
N = p·V / (R·T) and does not depend on the type of gas.

If you will the vessel under same conditions with a mixture of N(Ne) moles of Neon and N(Ar) moles of Argon, the total number of moles will be the same as in the first run:
N(Ne) + N(Ar) = N = 0.06073 mol
The total mass of gas is the number of moles times molar mass:
N(Ne) · M(Ne) + N(Ar) · M(Ar) = m'
which is measured as:
m' = 339.07g - 337.428g = 1.648g

Next introduce the mole fraction of Neon
x(Ne) = N(Ne)/N
from the equation for the number of moles you can easily derive:
x(Ar) = N(Ar)/N = 1 - x(Ne)
Divede the equation for the mass by N and get:
x(Ne) · M(Ne) + (1-x(Ne)) · M(Ar) = m'/N

Hence the mole fraction of Neon is given by:
x(Ne) = (M(Ar) - m'/N) = (M(Ar) - M(Ne))
= (39.948g/mol - 1.648g/0.0607mol) / (39.948g/mol - 20.1797g/mol)
= 64,8%

Answer 2

Data required.