A \(2.00-\mathrm{L}\) sample of \(\mathrm{O}_{2}(g)\) was collected over water at a total pressure of 785 torr and \(25^{\circ} \mathrm{C}\). When the \(\mathrm{O}_{2}(g)\) was dried (water vapor removed), the gas had a volume of \(1.94 \mathrm{~L}\) at \(25^{\circ} \mathrm{C}\) and 785 torr. Calculate the vapor pressure of water at \(25^{\circ} \mathrm{C}\).

Insert the calculated values of \(n_{wet}\) and \(n_{dry}\), and solve for the vapor pressure of water at \(25^{\circ} C\): \(n_{wet} = \frac{785\,\text{torr} \cdot 2.00\,\text{L}}{62.36\,\frac{\text{L}\cdot\text{torr}}{\text{mol}\cdot\text{K}} \cdot 298.15\, \text{K}} \approx 0.0840\, \text{mol}\) \(n_{dry} = \frac{785\,\text{torr} \cdot 1.94\,\text{L}}{62.36\,\frac{\text{L}\cdot\text{torr}}{\text{mol}\cdot\text{K}} \cdot 298.15\, \text{K}} \approx 0.0813\, \text{mol}\) \(n_{water} = n_{wet} - n_{dry} \approx 0.0840\, \text{mol} - 0.0813\, \text{mol} \approx 0.0027\, \text{mol}\) Now, calculate the partial pressure of water vapor: \(P_{H_2O} = 785\,\text{torr} \times \frac{0.0027\, \text{mol}}{0.0840\, \text{mol}} \approx 26\, \text{torr}\) The vapor pressure of water at \(25^{\circ} C\) is approximately \(26\,\text{torr}\).