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Chapter 6: Problem 68

An \(89.3 \mathrm{mL}\) sample of wet \(\mathrm{O}_{2}(\mathrm{g})\) is collected over water at \(21.3^{\circ} \mathrm{C}\) at a barometric pressure of \(756 \mathrm{mmHg}\) (vapor pressure of water at \(21.3^{\circ} \mathrm{C}=19 \mathrm{mmHg}\) ). (a) What is the partial pressure of \(\mathrm{O}_{2}(\mathrm{g})\) in the sample collected, in millimeters of mercury? (b) What is the volume percent \(\mathrm{O}_{2}\) in the gas collected? (c) How many grams of \(\mathrm{O}_{2}\) are present in the sample?

Short Answer

Expert verified
(a) The partial pressure of O2 in the gas sample is 737 mmHg. (b) The volume percent of O2 in the gas collected is 100%. (c) The mass of O2 present in the sample is approximately 0.1152 g.

Step by step solution

01

Calculate the partial pressure of O2

The gas sample collected is a mixture of O2 and water vapor, hence its total pressure is the sum of the partial pressures of the two gases. According to Dalton's law of partial pressures, the pressures of gases in a mixture add up to the total pressure. Therefore, the pressure of O2 can be calculated by subtracting the vapor pressure of water at the given temperature from the total pressure(Barometric pressure). \nPartial pressure of O2=Total pressure - Vapor pressure of water = 756 mmHg - 19 mmHg = 737 mmHg.
02

Calculation of volume percent of O2

The volume percent of oxygen in the gas mixture can be calculated by dividing the volume of oxygen gas by the total gas volume. Since the gas sample is wet, the total volume is taken as the volume of O2 plus the volume of the water vapor. However, the volume percentages of gases do not depend on the amounts, hence the volume of O2 is equal to total volume. Therefore, Volume% of O2 = (Volume of O2 / Total Volume) x 100 = (89.3 mL / 89.3 mL) x 100 = 100%.
03

Calculate the number of grams of O2

The partial pressure of O2 obtained in step 1 can be used to calculate the amount of oxygen in moles using the Ideal Gas Law (PV=nRT). Here, P is the partial pressure of the gas(737 mmHg or 0.97 atm), V is the volume (89.3 mL or 0.0893 L), R is the ideal gas constant (0.0821 L.atm/K.mol)' and T is the temperature in Kelvin (21.3°C = 294.45 K). First, rearrange the equation to solve for n (number of moles), n = PV / RT. Using these values, n = (0.97 atm x 0.0893 L) / (0.0821 L.atm/K.mol x 294.45 K) = 0.0036 mol. Finally, convert the moles into grams using the molar mass of oxygen (16 g/mol, as oxygen is diatomic, the molar mass is 32 g/mol). Mass = no. of moles x molar mass = 0.0036 mol x 32 g/mol = 0.1152 g.

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Most popular questions from this chapter

A \(0.156 \mathrm{g}\) sample of a magnesium-aluminum alloy dissolves completely in an excess of \(\mathrm{HCl}(\mathrm{aq}) .\) The liberated \(\mathrm{H}_{2}(\mathrm{g})\) is collected over water at \(5^{\circ} \mathrm{C}\) when the barometric pressure is 752 Torr. After the gas is collected, the water and gas gradually warm to the prevailing room temperature of \(23^{\circ} \mathrm{C} .\) The pressure of the collected gas is again equalized against the barometric pressure of 752 Torr, and its volume is found to be \(202 \mathrm{mL}\). What is the percent composition of the magnesium-aluminum alloy? (Vapor pressure of water: \(6.54 \mathrm{mmHg}\) at \(5^{\circ} \mathrm{C}\) and \(21.07 \mathrm{mmHg}\) at \(\left.23^{\circ} \mathrm{C}\right)\)

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