Question

A gaseous mixture of \(\mathrm{O}_{2}\) and \(\mathrm{Kr}\) has a density of \(1.104 \mathrm{~g} / \mathrm{L}\) at 435 torr and \(300 \mathrm{~K}\). What is the mole percent \(\mathrm{O}_{2}\) in the mixture?

Short answer

The mole percent of O₂ in the gaseous mixture of O₂ and Kr is 74.9%.

Step-by-step solution

Use the Ideal Gas Law to find the molar mass of the mixture

The Ideal Gas Law is given by \(PV = nRT\), where P is pressure, V is volume, n is moles, R is the ideal gas constant, and T is temperature. Since our pressure is given in torr, let's convert it to atm first by using the conversion factor \(1 \mathrm{~atm} = 760 \mathrm{~torr}\). This gives us: \(\text{Pressure} = \frac{435 \mathrm{~torr}}{760 \mathrm{~torr/atm}} = 0.5724 \mathrm{~atm}\) Now let's rearrange the Ideal Gas Law to find the molar mass of the mixture. Since \(PV = nRT\), we can express n in terms of M (molar mass) and mass (m) by the equation \(n = \frac{m}{M}\). Thus, \(PV = \frac{mRT}{M}\) and \(M = \frac{mRT}{PV}\). The density is given as 1.104 g/L, so \(m = 1.104 \mathrm{~g}\) and \(V = 1 \mathrm{~L}\). The Temperature is given as 300 K, and the ideal gas constant (R) is 0.0821 L.atm/mol.K. Then, we can find the molar mass (M) of the mixture: \(M = \frac{(1.104 \mathrm{~g})(0.0821 \mathrm{~L} \cdot \mathrm{atm}/ \mathrm{mol} \cdot \mathrm{K})(300 \mathrm{~K})}{(0.5724 \mathrm{~atm})(1 \mathrm{~L})}\)

Calculate the molar mass of the mixture

Now, let's plug in the values into the formula we derived above to find the molar mass of the mixture: \(M = \frac{(1.104)(0.0821)(300)}{(0.5724)(1)} = 51.16 \mathrm{~g/mol}\) So, the molar mass of the mixture is 51.16 g/mol.

Use mole fractions to find the mole percent of O₂ in the mixture

The molar mass of oxygen gas (O₂) is 32 g/mol, and the molar mass of krypton gas (Kr) is 83.8 g/mol. Let x be the mole fraction of O₂ in the mixture. The mole fraction of Kr in the mixture is then (1 - x). The molar mass of the mixture is given by the sum of the mole fractions times their respective molar masses: \(51.16 \mathrm{~g/mol} = x \times 32 \mathrm{~g/mol} + (1 - x) \times 83.8 \mathrm{~g/mol}\) Now let's solve this equation for x to find the mole fraction of O₂ in the mixture.

Solve for the mole fraction of O₂ in the mixture

To find x, we first rearrange the equation: \(51.16 = 32x + 83.8 - 83.8x\) Combining like terms, we get: \(51.16 = (32 - 83.8)x + 83.8\) Now, isolate x: \(x = \frac{51.16 - 83.8}{32 - 83.8}\) Calculating x, we get: \(x = 0.749\) Now that we have the mole fraction of O₂ in the mixture, we can find its mole percent by multiplying the mole fraction by 100. Mole percent of O₂ in the mixture = 0.749 × 100 = 74.9% The mole percent of O₂ in the gaseous mixture is 74.9%.