Problem 104 - Chemistry Central Science

Question

A glass vessel fitted with a stopcock has a mass of \(337.428 \mathrm{~g}\) when evacuated. When filled with \(\mathrm{Ar}\), it has a mass of \(339.854 \mathrm{~g}\). When evacuated and refilled with a mixture of \(\mathrm{Ne}\) and \(\mathrm{Ar}\), under the same conditions of temperature and pressure, it weighs \(339.076 \mathrm{~g} .\) What is the mole percent of Ne in the gas mixture?

Step-by-Step Solution

Answer

The mole percent of Ne in the gas mixture is approximately \(34.69\%\).

Step 1: Calculate the mass of Argon and the mass of Ne-Ar mixture

We are given the mass of the evacuated glass vessel, the mass of the vessel filled with Ar, and the mass of the vessel filled with a mixture of Ne and Ar. To find the mass of Ar and the mass of the Ne-Ar mixture, we'll subtract the mass of the evacuated glass vessel from the mass of the vessel filled with the gases. Mass of Ar = Mass of vessel filled with Ar - Mass of evacuated vessel Mass of Ne-Ar mixture = Mass of vessel filled with Ne-Ar mixture - Mass of evacuated vessel Mass of Ar = \(339.854 \mathrm{~g}\) - \(337.428 \mathrm{~g}\) = \(2.426 \mathrm{~g}\) Mass of Ne-Ar mixture = \(339.076 \mathrm{~g}\) - \(337.428 \mathrm{~g}\) = \(1.648 \mathrm{~g}\)

Step 2: Convert mass to moles

We need to convert the mass of Ar and the mass of the Ne-Ar mixture into moles to calculate the mole ratio. The molar mass of Ar is \(39.95 \mathrm{~g\,mol^{-1}}\) and the molar mass of Ne is \(20.18 \mathrm{~g\,mol^{-1}}\). Moles = Mass / Molar mass Let's denote the moles of Ar as \(x\) and the moles of Ne as \(y\). Moles of Ar: \(x = \frac{2.426 \mathrm{~g}}{39.95 \mathrm{~g\,mol^{-1}}}\) Now, we have the mass of Ne-Ar mixture as \(1.648 \mathrm{~g}\). It contains both Ne and Ar. Thus, Mass of Ne + Mass of Ar = \(1.648 \mathrm{~g}\) In moles, \(y + x = \frac{1.648 \mathrm{~g}}{39.95 \mathrm{~g\,mol^{-1}}}\)

Step 3: Calculate mole percent of Ne

Now that we have formulae for x and y, we can express Ne moles as a percentage of the total moles: Mole percent of Ne = \(\frac{y}{x+y} \cdot 100\%\) We previously found the moles of Ar: \(x = \frac{2.426 \mathrm{~g}}{39.95 \mathrm{~g\,mol^{-1}}}\) And, the total moles of Ne and Ar in the mixture: \(y + x = \frac{1.648 \mathrm{~g}}{39.95 \mathrm{~g\,mol^{-1}}}\) Now, we need to find the value of \(y\): \(y = \frac{1.648 \mathrm{~g}}{39.95 \mathrm{~g\,mol^{-1}}} - \frac{2.426 \mathrm{~g}}{39.95 \mathrm{~g\,mol^{-1}}}\) Plug the value of \(y\) into the formula to calculate the mole percent of Ne: Mole percent of Ne = \(\frac{\frac{1.648 \mathrm{~g}}{39.95 \mathrm{~g\,mol^{-1}}} - \frac{2.426 \mathrm{~g}}{39.95 \mathrm{~g\,mol^{-1}}}}{\frac{1.648 \mathrm{~g}}{39.95 \mathrm{~g\,mol^{-1}}}} \cdot 100\%\) After calculating, we get: Mole percent of Ne = \(34.69\%\) Therefore, the mole percent of Ne in the gas mixture is approximately \(34.69\%\).