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Chapter 6: Problem 140

A sample of \(\mathrm{O}_{2}(\mathrm{g})\) is collected over water at \(23^{\circ} \mathrm{C}\) and a barometric pressure of 751 Torr. The vapor pressure of water at \(23^{\circ} \mathrm{C}\) is \(21 \mathrm{mmHg}\). The partial pressure of \(\mathrm{O}_{2}(\mathrm{g})\) in the sample collected is (a) \(21 \mathrm{mmHg}_{i}\) (b) 751 Torr; \((\mathrm{c}) 0.96 \mathrm{atm} ;\) (d) \(1.02 \mathrm{atm}\).

Short Answer

Expert verified
The partial pressure of \(\mathrm{O}_{2}(\mathrm{g})\) in the sample collected is 0.96 atm (answer c).

Step by step solution

01

Conversion of units

Before moving ahead, ensure all the units are consistent. Here, both the barometric pressure is given in Torr and water vapour pressure is given in mmHg. Thankfully, 1 Torr is equal to 1 mmHg. Thus, no conversion is required for this problem.
02

Application of Dalton's Law of Partial Pressures

As per Dalton's Law, the total pressure is the sum of the individual partial pressures.In this scenario, total pressure (P total) is the barometric pressure, and this consists of two partial pressures - the pressure from O2 gas (P O2) and the vapor pressure of water (P water). We know, P total = P O2 + P water. Thus, re-arranging, we get P O2 = P total - P water. Given, P total = 751 mmHg and P water = 21 mmHg, substituting these values into the equation gives P O2 = 751 mmHg - 21 mmHg = 730 mmHg.
03

Convert mmHg to atm

Finally, we want to convert the pressure of O2 from mmHg to atm, as the options are given in atm. We use the conversion factor that 1 atm equals 760 mmHg. Thus, to get the pressure in atm, we divide the pressure in mmHg by 760. Hence, P O2 in atm = 730 mmHg / 760 mmHg/atm = 0.96 atm.

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