Patients undergoing an upper gastrointestinal tract laboratory test are typically given an X-ray contrast agent that aids with the radiologic imaging of the anatomy. One such contrast agent is sodium diatrizoate, a nonvolatile water-soluble compound. A 0.378 -m solution is prepared by dissolving 38.4 \(\mathrm{g}\) sodium diatrizoate (NaDTZ) in \(1.60 \times 10^{2} \mathrm{mL}\) water at \(31.2^{\circ} \mathrm{C}\) (the density of water at $31.2^{\circ} \mathrm{C}\( is 0.995 \)\mathrm{g} / \mathrm{cm}^{3} ) .$ What is the molar mass of sodium diatrizoate? What is the vapor pressure of this solution if the vapor pressure of pure water at \(31.2^{\circ} \mathrm{C}\) is 34.1 torr?

First, we need to find the mass of water in the solution. We have the volume and density of water at 31.2°C. We can multiply the volume by the density to get the mass of water. $$ mass_{water} = volume_{water} \times density_{water} $$ Plug in the given values: $$ mass_{water} = (1.60 \times 10^2 \mathrm{mL}) \times (0.995 \mathrm{g/mL}) $$ Compute for the mass of water: $$ mass_{water} = 159.2 \mathrm{g} $$

To find the molality, we will use the formula: $$ molality = \frac{moles_{solute}}{mass_{solvent}(kg)} $$ We are given the molality of the solution (0.378 mol/kg) and the mass of water (solvent) in the solution (159.2g or 0.1592kg). Solve for moles of NaDTZ: $$ moles_{NaDTZ} = molality \times mass_{solvent}(kg) $$ Plug in the given values: $$ moles_{NaDTZ} = 0.378 \mathrm{mol/kg} \times 0.1592 \mathrm{kg} $$ Compute for the moles of NaDTZ: $$ moles_{NaDTZ} = 0.0602 \mathrm{mol} $$

To find the molar mass of NaDTZ, we will use the formula: $$ molar \, mass_{NaDTZ} = \frac{mass_{NaDTZ}}{moles_{NaDTZ}} $$ We are given the mass of NaDTZ (38.4g) and we have calculated the moles of NaDTZ (0.0602mol). Plug in these values: $$ molar \, mass_{NaDTZ} = \frac{38.4 \mathrm{g}}{0.0602 \mathrm{mol}} $$ Compute for the molar mass of NaDTZ: $$ molar \, mass_{NaDTZ} = 638 \mathrm{g/mol} $$

To apply Raoult's Law, we first need to find the mole fraction of NaDTZ in the solution. The mole fraction can be calculated using the following formula: $$ \chi_{NaDTZ} = \frac{moles_{NaDTZ}}{moles_{NaDTZ} + moles_{water}} $$ Water has a molar mass of 18.015 g/mol. Therefore, we can compute the moles of water in the solution: $$ moles_{water} = \frac{mass_{water}}{molar \, mass_{water}} = \frac{159.2 \mathrm{g}}{18.015 \mathrm{g/mol}} = 8.84 \mathrm{mol} $$ Now compute the mole fraction of NaDTZ in the solution: $$ \chi_{NaDTZ} = \frac{0.0602 \mathrm{mol}}{0.0602 \mathrm{mol} + 8.84 \mathrm{mol}} $$ Compute for the mole fraction of NaDTZ: $$ \chi_{NaDTZ} = 0.00677 $$

Raoult's Law states that the vapor pressure of the solution is the product of the mole fraction of the solute and the vapor pressure of the pure solute. In this case, the solute is NaDTZ, and the solvent is water. Since NaDTZ is a non-volatile solute, its vapor pressure is considered to be 0. Therefore, the vapor pressure of the solution is given by: $$ P_{solution} = \chi_{water} \times P_{water} $$ Here, \( \chi_{water} = 1 - \chi_{NaDTZ} \), and we are given the vapor pressure of pure water at 31.2°C (34.1 torr). Compute the mole fraction of water and the vapor pressure of the solution: $$ \chi_{water} = 1 - \chi_{NaDTZ} = 1 - 0.00677 = 0.993 $$ $$ P_{solution} = \chi_{water} \times P_{water} = 0.993 \times 34.1 \mathrm{torr} $$ Compute for the vapor pressure of the solution: $$ P_{solution} = 33.9 \mathrm{torr} $$ To summarize: - The molar mass of sodium diatrizoate (NaDTZ) is 638 g/mol. - The vapor pressure of the NaDTZ solution at 31.2°C is 33.9 torr.