Un recipiente de vidrio provisto con una llave de paso tiene...

Un recipiente de vidrio provisto con una llave de paso tiene una masa de 337.428 g cuando está evacuado. Si este recipiente se llena con $\mathrm{Ar}$, tiene una masa de $339.854 \mathrm{~g}$. Si el recipiente se evacua y se vuelve a llenar con una mezcla de Ne y Ar, en las mismas condiciones de temperatura y presión, tiene una masa de $339.076 \mathrm{~g}$. Calcule el porcentaje en mol de Ne en la mezcla de gases.

Key Concepts

Ideal Gas Law

This law relates the pressure, volume, temperature, and number of moles of a gas. In problems where gases are at the same conditions (constant temperature and pressure) in a fixed volume, it implies that the number of moles is directly proportional to the amount of gas present. This concept underpins the interpretation of mass measurements in terms of moles in gas mixtures.

Mass-to-Mole Conversion

This concept involves converting a known mass of a substance into the number of moles using its molar mass. When comparing different gases, knowing their molar masses allows one to determine the moles present from the measured mass, which is crucial for analyzing the composition of a gas mixture.

Mole Fraction

The mole fraction expresses the ratio of the number of moles of one component to the total number of moles in the mixture. In gas mixtures, mole fractions are used to describe the composition irrespective of the individual gas masses and are central to determining the percent composition in moles.

Transcript

00:01 Well, let's start by finding the mass of our argon.

00:05 Okay, that's going to be the mass of the vessel with the argon in it, minus the empty vessel, which is 337 .48.

00:19 So we can see that our mass of argon is going to be 2 .46 grams.

00:27 Well, let's go ahead and change that to moles, using the molar mass, and we'll see that that is going to be 0 .0 .06 -073 moles of argon.

00:43 So if that many moles of argon fit in there, this is also the total moles that can fit in that flask regardless of what that gas is going to be.

00:54 So when we have it as argon and neon, it's still going to have that total moles.

01:00 So let's say the second time we fill it, it's 339.

01:08 0 .076.

01:10 Again, let's subtract the empty mass.

01:16 And we'll see that we have a total of 1 .648 grams of argon and neon.

01:24 So i'm going to let x equal the mass of argon in grams.

01:33 And i'm going to let 1 .648 minus x equals the mass of my neon, also in grams.

01:44 So we know that the mass of the argon plus the mass of the neon has to equal 1 .648.

01:51 Okay, and we also, so i'm going to take x, and i'm going to divide it by the molar mass of my argon.

02:05 So now what i have is moles of argon, and i'm going to add that 1 .648 minus x grams of neon.

02:18 I'm going to divide by its molar mass.

02:21 So this would be the moles of neon, right? grams over grams per mole.

02:28 So this has to equal our total moles, which we said was 0 .0673.

02:36 So this whole thing will equal 0 .0673...

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