A mixture of \(\mathrm{H}_{2}(\mathrm{g})\) and \(\mathrm{O}_{2}(\mathrm{g})\) is prepared by electrolyzing \(1.32 \mathrm{g}\) water, and the mixture of gases is collected over water at \(30^{\circ} \mathrm{C}\) and \(748 \mathrm{mmHg} .\) The volume of "wet" gas obtained is 2.90 L. What must be the vapor pressure of water at \(30^{\circ} \mathrm{C} ?\) $$2 \mathrm{H}_{2} \mathrm{O}(1) \stackrel{\text { electrolysis }}{\longrightarrow} 2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})$$

The electrolysis of water generates hydrogen and oxygen gases. The equation tells us that from 2 mole of water, 2 moles of hydrogen gas and 1 mole of oxygen gas are produced.

To calculate the moles of water that undergo electrolysis, use the formula: \n moles = mass / molecular weight. \nThe molecular weight of water is 18g/mol and given mass is 1.32g. \nThe calculation gives us \(\frac{1.32g}{18g/mol} = 0.073 mol\)

From the balanced chemical equation in the question, it can be inferred that the electrolysis of \(2\) moles of water (\(H_{2}O\)) results in \(2\) moles of hydrogen gas (\(H_{2}\)) and \(1\) mole of oxygen gas (\(O_{2}\)). Using this stoichiometric ratio, find out the total moles of gases produced.\n Total moles of gas = Moles of \(H_2\) + Moles of \(O_2\) = \(2\times 0.073\) (from hydrogen) + \(0.073\) (from oxygen) = \(0.219 mol\)

The ideal gas law equation is: \(P = \frac {nRT}{V}\), where P is pressure, n is moles, R is gas constant, T is temperature and V is volume.\n We are given that the total volume of wet gas produced is \(2.90 L\) and we have calculated the total moles to be \(0.219 mol\).\n The value of R (gas constant) is \(0.0821 \frac {L.atm}{K.mol}\).\n The temperature given is already in Celsius and it needs to be converted to Kelvin (by adding 273 to the Celsius value), resulting in \(30 + 273 = 303K\).\n Substituting the calculated and given values into the ideal gas law equation gives us:\n \(P = \frac {nRT}{V} = \frac {(0.219 mol) * (0.0821 \frac {L.atm}{K.mol}) * (303K)}{2.90L}\).\n This gives the total pressure of collected gases to be \(0.61 atm\)

Here we need to remember Dalton's Law of Partial Pressures which states that the total pressure of a gaseous mixture is equal to the sum of the partial pressures of each individual gas.\n Thus, the pressure of the pure gases collected (which we have calculated) plus the vapor pressure of water at \(30^{\circ}C\) should equal the total atmospheric pressure given in the problem. The total atmospheric pressure is given in mmHg, for our calculations it is more convenient to have it in atmospheres. The conversion is \(1 atm = 760 mmHg\), so \(748 mmHg = 0.984 atm\).\n The vapor pressure of water at \(30^{\circ}C\) can be calculated by subtracting the total gas pressure from the atmospheric pressure:\n Vapor pressure of water = Total atmospheric pressure - total gas pressure = \(0.984 atm - 0.61 atm = 0.374 atm\)