Which of the following will have the lowest total vapor pressure at \(25^{\circ} \mathrm{C} ?\) a. pure water (vapor pressure \(=23.8\) torr at \(25^{\circ} \mathrm{C} )\) b. a solution of glucose in water with $\chi_{\mathrm{CH}_{2} \mathrm{O}_{6}}=0.01$ c. a solution of sodium chloride in water with \(\chi_{\mathrm{NaCl}}=0.01\) d. a solution of methanol in water with \(\chi_{\mathrm{CH}, 0 \mathrm{H}}=0.2\) (Consider the vapor pressure of both methanol \([143 \text { torr at }\) \(25^{\circ} \mathrm{C} ]\) and water.

For pure water, the vapor pressure is already given as \(23.8\) torr. So, no further calculation is needed for this case.

Glucose is a non-volatile solute, which means it doesn't contribute to the vapor pressure. Using Raoult's law, we can find the vapor pressure of water in this solution: \[P_{\text{H}_2\text{O}} = \chi_{\text{H}_2\text{O}} * P_{\text{pure H}_2\text{O}}\] Since mole fraction of glucose (\(\chi_{\mathrm{CH}_{2}\mathrm{O}_{6}}\)) is \(0.01\), mole fraction of water is \(1 - 0.01 = 0.99\). Therefore, \[P_{\text{H}_2\text{O}} = 0.99 * 23.8 \mathrm{torr} = 23.562 \mathrm{torr}\]

Since NaCl is an ionic compound, its dissolution into water will increase the number of solute particles, forming 2 ions (Na\(^+\) and Cl\(^-\)) for each NaCl formula unit. This will affect the vapor pressure, which does not follow Raoult's law ideally: \[P_{\text{NaCl}} = i * \chi_{\text{H}_2\text{O}} * P_{\text{pure H}_2\text{O}}\] where \(i\) is the van't Hoff factor, in this case \(i=2\). For this case, \(\chi_{\mathrm{NaCl}}=0.01\), thus mole fraction of water is \(1-0.01=0.99\). Then, \[P_{\text{H}_2\text{O}} = 2 * 0.99 * 23.8 \mathrm{torr} = 47.124 \mathrm{torr}\]

For the methanol-water solution, we have two volatile components, which means we need to take into account both components' vapor pressure. Applying Raoult's law for each component: \[P_{\text{H}_2\text{O}} = \chi_{\text{H}_2\text{O}} * P_{\text{pure H}_2\text{O}}\] \[P_{\text{CH}_3\text{OH}} = \chi_{\text{CH}_3\text{OH}} * P_{\text{pure CH}_3\text{OH}}\] Here, \(\chi_{\mathrm{CH}, 0 \mathrm{H}}=0.2\), thus mole fraction of water is \(1-0.2=0.8\). Then, \[P_{\text{H}_2\text{O}} = 0.8 * 23.8 \mathrm{torr} = 19.04 \mathrm{torr}\] \[P_{\text{CH}_3\text{OH}} = 0.2 * 143 \mathrm{torr}= 28.6 \mathrm{torr}\] The total vapor pressure of the methanol solution is the sum of the partial pressures of water and methanol. \[P_{\text{total}} = P_{\text{H}_2\text{O}} + P_{\text{CH}_3\text{OH}} = 19.04 \mathrm{torr} + 28.6 \mathrm{torr} = 47.64 \mathrm{torr}\]

Now, let's compare the vapor pressures obtained for each case: Pure water: \(23.8 \mathrm{torr}\) Glucose solution: \(23.562 \mathrm{torr}\) NaCl solution: \(47.124 \mathrm{torr}\) Methanol solution: \(47.64 \mathrm{torr}\) The solution with the lowest total vapor pressure is the glucose solution (b).