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Chapter 6: Problem 102

A gaseous mixture of He and \(\mathrm{O}_{2}\) has a density of \(0.518 \mathrm{g} / \mathrm{L}\) at \(25^{\circ} \mathrm{C}\) and \(721 \mathrm{mm} \mathrm{Hg} .\) What is the mass percent He in the mixture?

Short Answer

Expert verified
The mass percent of \(He\) in the mixture is approximately 20\%.

Step by step solution

01

Convert pressure to atm

First, convert the pressure to the standard units of atmospheres (atm). The conversion factor is 1 atm = 760 mmHg, so pressure = \(721 \, \mathrm{mmHg} \times \frac{1 \, \mathrm{atm}}{760 \, \mathrm{mmHg}} = 0.95 \, \mathrm{atm}\).
02

Convert temperature to Kelvin

Second, convert the temperature from Celsius to the standard units of Kelvin (K). The conversion factor is \(K = ^{\circ}C + 273.15\), so temperature = \(25^{\circ}C + 273.15 = 298.15K\).
03

Use the ideal gas equation to find moles of gas

Next, use the ideal gas equation \(PV = nRT\) to find the number of moles of gas in one liter. Rearrange the equation for n: \(n = \frac{PV}{RT}\). Using R = 0.0821 L atm/mol K, the number of moles of gas per liter is \((0.95 \, \mathrm{atm} \times 1 \, \mathrm{L}) / (0.0821 \, \mathrm{L} \, \mathrm{atm}/\mathrm{mol} \, \mathrm{K} \times 298.15 \, \mathrm{K}) = 0.0385 \, \mathrm{mol/L}\).
04

Calculate Total mass and Individual masses

The density of the gas mixture is given as 0.518 g/L so the total mass is given by density x volume = \(0.518 \, \mathrm{g}/\mathrm{L} \times 1 \, \mathrm{L} = 0.518g\). Also, The molar mass of oxygen (O2) is 32 g/mol and helium is 4 g/mol. We use the total number of moles (0.0385 mol) to calculate the mass of helium and oxygen individually. We then use these masses to calculate the percentage mass contribution of helium.
05

Solve for the mass percent of He

Finally, calculate the mass percent of \(He\), which is the mass of \(He\) divided by the total mass of the gas mixture times 100. Let x represent the molar fraction of \(He\) in the gas mixture, then the fraction of \(O2\) is \(1-x\). Set up an equation: \(x \times 4g/mol \times 0.0385 mol + (1-x) \times 32 g/mol \times 0.0385 mol = 0.518 g\). Solve for x and calculate the mass percent, it gives approximately 20\%.

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Most popular questions from this chapter

Hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2},\) is used to disinfect contact lenses. How many milliliters of \(\mathrm{O}_{2}(\mathrm{g})\) at \(22^{\circ} \mathrm{C}\) and \(752 \mathrm{mmHg}\) can be liberated from \(10.0 \mathrm{mL}\) of an aqueous solution containing \(3.00 \% \mathrm{H}_{2} \mathrm{O}_{2}\) by mass? The density of the aqueous solution of \(\mathrm{H}_{2} \mathrm{O}_{2}\) is \(1.01 \mathrm{g} / \mathrm{mL}\) $$2 \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(1)+\mathrm{O}_{2}(\mathrm{g})$$

If 3.0 L of oxygen gas at \(177^{\circ} \mathrm{C}\) is cooled at constant pressure until the volume becomes \(1.50 \mathrm{L}\), then what is the final temperature?

According to the CRC Handbook of Chemistry and Physics (83rd ed.), the molar volume of \(\mathrm{O}_{2}(\mathrm{g})\) is \(0.2168 \mathrm{Lmol}^{-1}\) at \(280 \mathrm{K}\) and \(10 \mathrm{MPa}\). (Note: \(1 \mathrm{MPa}=\) \(\left.1 \times 10^{6} \mathrm{Pa} .\right)\)(a) Use the van der Waals equation to calculate the pressure of one mole of \(\mathrm{O}_{2}(\mathrm{g})\) at \(280 \mathrm{K}\) if the volume is 0.2168 L. What is the \% error in the calculated pressure? The van der Waals constants are \(a=1.382 \mathrm{L}^{2}\) bar \(\mathrm{mol}^{-2}\) and \(b=0.0319 \mathrm{L} \mathrm{mol}^{-1}\) (b) Use the ideal gas equation to calculate the volume of one mole of \(\mathrm{O}_{2}(\mathrm{g})\) at \(280 \mathrm{K}\) and \(10 \mathrm{MPa}\). What is the \% error in the calculated volume?

At what temperature is the molar volume of an ideal gas equal to \(22.4 \mathrm{L},\) if the pressure of the gas is \(2.5 \mathrm{atm} ?\)

Convert each pressure to an equivalent pressure (a) 736 mmHg; (b) 0.776 bar; in atmospheres. (c) 892 Torr; (d) 225 kPa.
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