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Chapter 6: Problem 69

A sample of \(\mathrm{O}_{2}(\mathrm{g})\) is collected over water at \(24^{\circ} \mathrm{C}\) The volume of gas is 1.16 L. In a subsequent experiment, it is determined that the mass of \(\mathrm{O}_{2}\) present is 1.46 g. What must have been the barometric pressure at the time the gas was collected? (Vapor pressure of water \(=22.4 \text { Torr. })\)

Short Answer

Expert verified
The barometric pressure at the time the gas was collected was 1.15 atm.

Step by step solution

01

Identification of Moles of the Oxygen Gas

Use the molar mass of oxygen to identify the number of moles in the given mass. The molar mass of \(O_2\) is \(32.00 \, g/mol\). Thus, the number of moles \(n\) of \(O_2\) is given by the formula: \(n = \frac{mass}{molar \,mass} = \frac{1.46 \, g}{32.00 \, g/mol} = 0.046 \, mol\).
02

Calculation of Total Pressure using the Ideal Gas Law

We can use the Ideal Gas law \(PV = nRT\), where \(P\) is the total pressure, \(V\) is the volume in liters, \(n\) is the number of moles, \(R\) is the gas constant (\(0.0821 \, L \cdot atm/(mol \cdot K)\)), and \(T\) is the temperature in Kelvin. First, convert the temperature from Celsius to Kelvin using the formula \(K = °C + 273.15\) which gives \(T = 24 + 273.15 = 297.15 \, K\). Then, substitute the appropriate values to find \(P\): \(P = \frac{nRT}{V} = \frac{0.046 \, mol \cdot 0.0821 \, L \cdot atm/(mol \cdot K) \cdot 297.15 \, K}{1.16 \, L} = 1.12 \, atm\).
03

Applying Dalton's Law of Partial Pressure

Dalton's law of partial pressures states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases. In this case, that means the total pressure \(P_{total}\) is equal to the pressure of oxygen \(P_{O2}\) plus the vapor pressure of water \(P_{H2O}\). Since we're looking for the barometric pressure, which is the same as the total pressure, we rearrange the equation to be: \(P_{total} = P_{O2} + P_{H2O}\) or \(P_{Barometric} = P_{O2} + P_{H2O}\). The given water pressure is \(22.4 \, Torr = 0.0295 \, atm\). Therefore, the barometric pressure is \(1.12 \, atm + 0.0295 \, atm = 1.15 \, atm\).

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Most popular questions from this chapter

A sample of \(\mathrm{N}_{2}(\mathrm{g})\) effuses through a tiny hole in \(38 \mathrm{s}\) What must be the molar mass of a gas that requires \(64 \mathrm{s}\) to effuse under identical conditions?

Under which conditions is \(\mathrm{Cl}_{2}\) most likely to behave like an ideal gas? Explain. (a) \(100^{\circ} \mathrm{C}\) and \(10.0 \mathrm{atm}\) (b) \(0^{\circ} \mathrm{C}\) and 0.50 atm; \((\mathrm{c}) 200^{\circ} \mathrm{C}\) and \(0.50 \mathrm{atm}\) (d) \(400^{\circ} \mathrm{C}\) and \(10.0 \mathrm{atm}\).

To establish a pressure of 2.00 atm in a 2.24 L cylinder containing \(1.60 \mathrm{g} \mathrm{O}_{2}(\mathrm{g})\) at \(0^{\circ} \mathrm{C},\) (a) add \(1.60 \mathrm{g} \mathrm{O}_{2} ;(\mathrm{b})\) add \(0.60 \mathrm{g} \mathrm{He}(\mathrm{g}) ;(\mathrm{c})\) add \(2.00 \mathrm{g} \mathrm{He}(\mathrm{g})\) (d) release \(0.80 \mathrm{g} \mathrm{O}_{2}(\mathrm{g})\)

The equation \(d / P=M / R T,\) which can be derived from equation \((6.14),\) suggests that the ratio of the density \((d)\) to pressure (P) of a gas at constant temperature should be a constant. The gas density data at the end of this question were obtained for \(\mathrm{O}_{2}(\mathrm{g})\) at various pressures at \(273.15 \mathrm{K}\) (a) Calculate values of \(d / P,\) and with a graph or by other means determine the ideal value of the term \(d / P\) for \(\mathrm{O}_{2}(\mathrm{g})\) at \(273.15 \mathrm{K}\) [Hint: The ideal value is that associated with a perfect (ideal) gas.] (b) Use the value of \(d / P\) from part (a) to calculate a precise value for the atomic mass of oxygen, and compare this value with that listed on the inside front cover. $$\begin{array}{lllll} P, \mathrm{mmHg}: & 760.00 & 570.00 & 380.00 & 190.00 \\ d, \mathrm{g} / \mathrm{L}: & 1.428962 & 1.071485 & 0.714154 & 0.356985 \end{array}$$

A 3.05 g sample of \(\mathrm{NH}_{4} \mathrm{NO}_{3}(\mathrm{s})\) is introduced into an evacuated 2.18 L flask and then heated to \(250^{\circ} \mathrm{C}\).What is the total gas pressure, in atmospheres, in the flask at \(250^{\circ} \mathrm{C}\) when the \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) has completely decomposed? $$\mathrm{NH}_{4} \mathrm{NO}_{3}(\mathrm{s}) \longrightarrow \mathrm{N}_{2} \mathrm{O}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$
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