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Chapter 10: Problem 112

A gaseous mixture of \(\mathrm{O}_{2}\) and \(\mathrm{Kr}\) has a density of \(1.104 \mathrm{~g} / \mathrm{L}\) at 355 torr and \(400 \mathrm{~K}\). What is the mole percent \(\mathrm{O}_{2}\) in the mixture?

Short Answer

Expert verified
The mole percent of Oxygen (\(\mathrm{O}_2\)) in the gaseous mixture is 75.5%.

Step by step solution

01

1. Write the Ideal Gas Law formula in terms of density and molar mass

The Ideal Gas Law is given by: \(PV = nRT\) where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature. We can write the Ideal Gas Law in terms of molar mass and density as follows: \(P = (\rho \frac{RT}{M})\) where P is the pressure, ρ is the density, R is the gas constant, T is the temperature, and M is the molar mass of the gas.
02

2. Calculate the molar mass of the mixture given its density, pressure, and temperature

Given the density (ρ) as 1.104 g/L, pressure (P) as 355 torr, and temperature (T) as 400 K, we can calculate the molar mass (M) of the mixture. Make sure to convert the pressure from torr to atm by dividing it by 760. \(P(atm) = \frac{355 \ torr}{760 \ torr/atm} = 0.4671 \ atm\) Now, we can plug in the given values into the Ideal Gas Law and solve for M: \(0.4671 \ atm = (\frac{1.104 \ g/L \cdot 0.0821 \ L \cdot atm/mol \cdot K \cdot 400 \ K}{M})\) Solving for M, we get: \(M = 9.90 \ g/mol\)
03

3. Write the expression for molar mass in terms of a mole fraction

Let x be the mole fraction of Oxygen in the mixture. Then, the mole fraction of Krypton is (1-x). The expression for the molar mass of the mixture is given by: \(M = x \cdot M_{O_2} + (1 - x) \cdot M_{Kr}\) where \(M_{O_2}\) and \(M_{Kr}\) are the molar masses of Oxygen and Krypton, respectively. The molar masses of Oxygen and Krypton are 32 g/mol and 83.798 g/mol, respectively.
04

4. Calculate the mole fraction of Oxygen in the mixture

Now we can set up the equation with the value of M that we found in Step 2 and the molar masses of Oxygen and Krypton: \(9.90 \ g/mol = x \cdot 32 \ g/mol + (1 - x) \cdot 83.798 \ g/mol\) Solving this equation for x, we get the mole fraction of Oxygen: \(x = 0.755\)
05

5. Calculate the mole percent of Oxygen in the mixture

Finally, we can convert the mole fraction of Oxygen to a mole percent by multiplying it by 100: Mole percent O2 = x * 100 = 0.755 * 100 = 75.5% Therefore, the mole percent of Oxygen in the gaseous mixture is 75.5%.

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Most popular questions from this chapter

The physical fitness of athletes is measured by " \(V_{\mathrm{O}_{2}}\) max," which is the maximum volume of oxygen consumed by an individual during incremental exercise (for example, on a treadmill). An average male has a \(V_{\mathrm{O}_{2}}\) max of \(45 \mathrm{~mL} \mathrm{O}_{2} / \mathrm{kg}\) body mass \(/ \mathrm{min}\), but a world-class male athlete can have a \(V_{\mathrm{O}_{2}}\) max reading of \(88.0 \mathrm{~mL} \mathrm{O}_{2} / \mathrm{kg}\) body mass \(/ \mathrm{min.}\) (a) Calculate the volume of oxygen, in mL, consumed in 1 hr by an average man who weighs 185 lbs and has a \(V_{\mathrm{O}_{2}}\) max reading of 47.5 \(\mathrm{mL} \mathrm{O}_{2} / \mathrm{kg}\) body mass \(/ \mathrm{min}\) (b) If this man lost \(20 \mathrm{lb}\), exercised, and increased his \(V_{\mathrm{O}_{2}}\) max to \(65.0 \mathrm{~mL} \mathrm{O}_{2} / \mathrm{kg}\) body mass \(/ \mathrm{min}\), how many mL of oxygen would he consume in \(1 \mathrm{hr}\) ?

Large amounts of nitrogen gas are used in the manufacture of ammonia, principally for use in fertilizers. Suppose \(120.00 \mathrm{~kg}\) of \(\mathrm{N}_{2}(g)\) is stored in a 1100.0 - \(\mathrm{L}\) metal cylinder at \(280^{\circ} \mathrm{C}\). (a) Calculate the pressure of the gas, assuming ideal-gas behavior. (b) By using data in Table \(10.3,\) calculate the pressure of the gas according to the van der Waals equation. (c) Under the conditions of this problem, which correction dominates, the one for finite volume of gas molecules or the one for attractive interactions?

On a single plot, qualitatively sketch the distribution of molecular speeds for \((\mathbf{a}) \operatorname{Kr}(g)\) at \(-50^{\circ} \mathrm{C},(\mathbf{b}) \operatorname{Kr}(g)\) at \(0^{\circ} \mathrm{C},(\mathbf{c}) \operatorname{Ar}(g)\) at \(0{ }^{\circ} \mathrm{C}\). [Section \(\left.10.7\right]\)

Does the effect of intermolecular attraction on the properties of a gas become more significant or less significant if (a) the gas is compressed to a smaller volume at constant temperature; (b) the temperature of the gas is increased at constant volume?

Calculate each of the following quantities for an ideal gas: (a) the volume of the gas, in liters, if \(1.50 \mathrm{~mol}\) has a pressure of 1.25 atm at a temperature of \(-6^{\circ} \mathrm{C} ;(\mathbf{b})\) the absolute temperature of the gas at which \(3.33 \times 10^{-3}\) mol occupies \(478 \mathrm{~mL}\) at 750 torr; \((\mathbf{c})\) the pressure, in atmospheres, if \(0.00245 \mathrm{~mol}\) occupies \(413 \mathrm{~mL}\) at \(138{ }^{\circ} \mathrm{C} ;(\mathbf{d})\) the quantity of gas, in moles, if \(126.5 \mathrm{~L}\) at \(54^{\circ} \mathrm{C}\) has a pressure of \(11.25 \mathrm{kPa}\).
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