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Chapter 13: Problem 46

At 1.00 atm, the solubility of \(\mathrm{O}_{2}\) in water is \(2.18 \times 10^{-3} \mathrm{M}\) at \(0^{\circ} \mathrm{C}\) and \(1.26 \times 10^{-3} \mathrm{M}\) at \(25^{\circ} \mathrm{C}\) What volume of \(\mathrm{O}_{2}(\mathrm{g}),\) measured at \(25^{\circ} \mathrm{C}\) and \(1.00 \mathrm{atm},\) is expelled when \(515 \mathrm{mL}\) of water saturated with \(\mathrm{O}_{2}\) is heated from 0 to \(25^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
11.6 mL

Step by step solution

01

Understanding the solubility

Solubility at \(0^{\circ} \mathrm{C}\) is \(2.18 \times 10^{-3}\) M and at \(25^{\circ} \mathrm{C}\) it is \(1.26 \times 10^{-3} \mathrm{M}\). The molarity difference implies how much \(\mathrm{O}_{2}(\mathrm{g})\) is expelled when the temperature rises.
02

Calculate the change in molarity

The change in solubility (change in molarity) is \(2.18 \times 10^{-3} \mathrm{M} - 1.26 \times 10^{-3} \mathrm{M} = 0.92 \times 10^{-3}\) M or \(9.2 \times 10^{-4} \mathrm{M}\). This is the amount of \(\mathrm{O}_{2}(\mathrm{g})\) that is expelled from each liter of water when it is heated from \(0^{\circ} \mathrm{C}\) to \(25^{\circ} \mathrm{C}\).
03

Find the moles of \(\mathrm{O}_{2}(\mathrm{g})\) expelled

Use the change in molarity and the volume of the water to find the moles of \(\mathrm{O}_{2}(\mathrm{g})\) expelled. Use the formula for molarity \(M = \frac{n}{V}\), where V is in liters. Solve for n (moles of gas): \(n = M \times V = 9.2 \times 10^{-4} \mathrm{M} \times 0.515 \mathrm{L} = 4.74 \times 10^{-4} \mathrm{mol}\)
04

Calculate the volume of gas expelled

Under the conditions of \(25^{\circ}C\) and \(1.00 atm\), 1 mole of any gas occupies 24.5 L. Therefore, the volume of the \(\mathrm{O}_{2}\) gas expelled when the solution is heated is: \(V = 4.74 \times 10^{-4} \mathrm{mol} \times 24.5 \mathrm{L/mol} = 0.0116 \mathrm{L} = 11.6 \mathrm{mL}\)

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