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 \(0.378 \mathrm{~m}\) 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 \(\left.0.995 \mathrm{~g} / \mathrm{mL}\right)\). 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?

Using the given molality and mass of sodium diatrizoate, we can find the number of moles of NaDTZ: \(Malality = \frac{Number ~of~ moles ~of~ NaDTZ}{Mass ~of~ solvent ~in~ kg}\) Rearranging the equation: \(Number ~of~ moles ~of~ NaDTZ = Molality \times Mass ~of~ solvent ~in~ kg\) Given the volume of water (160 mL) and the density (0.995 g/mL) at 31.2 °C, we can find the mass of solvent (water) in grams and then convert it to kilograms: \(Mass ~of~ water = Volume ~of~ water \times Density ~of~ water\) \(Mass ~of~ water = 160 mL \times 0.995 \frac{g}{mL} = 159.2 g\) Converting grams to kilograms: \(Mass ~of~ water = \frac{159.2 g}{1000} = 0.1592 kg\) Now, let's find the number of moles of NaDTZ in the solution using the given molality (0.378 m): \(Number ~of~ moles ~of~ NaDTZ = 0.378 m \times 0.1592 kg = 0.0602 ~moles\)

Now, we can find the molar mass of NaDTZ using the mass and number of moles: \(Molar ~mass ~of ~NaDTZ = \frac{Mass ~of ~NaDTZ}{Number ~of ~moles ~of ~NaDTZ}\) Given the mass of NaDTZ (38.4 g), we can calculate the molar mass: \(Molar ~mass ~of ~NaDTZ = \frac{38.4 g}{0.0602 ~moles} = 638 \frac{g}{mol}\) So, the molar mass of sodium diatrizoate is 638 g/mol.

To find the vapor pressure of the solution, we first need to determine the mole fraction of water in the solution. The mole fraction of a component is the ratio of the number of moles of that component to the total number of moles in the solution. We know the number of moles of NaDTZ (0.0602 moles) and the mass of water (159.2 g). We can find the number of moles of water by dividing the mass by the molar mass of water (18.015 g/mol): \(Number ~of ~moles ~of ~water = \frac{Mass ~of ~water}{Molar ~mass ~of ~water} = \frac{159.2 g}{18.015 \frac{g}{mol}} = 8.842 ~moles\) Now, we can find the mole fraction of water (X_water) in the solution: \(X_{water} = \frac{Number ~of ~moles ~of ~water}{Total ~number ~of ~moles ~in ~solution} = \frac{8.842 ~moles}{8.842 ~moles + 0.0602 ~moles} = 0.993\)

Now that we have the mole fraction of water, we can apply Raoult's law to find the vapor pressure of the solution: \(Vapor ~pressure ~of ~solution = X_{water} \times Vapor ~pressure ~of ~pure ~water\) Given the vapor pressure of pure water at 31.2 °C (34.1 torr), we can calculate the vapor pressure of the solution: \(Vapor ~pressure ~of ~solution = 0.993 \times 34.1 ~torr = 33.9 ~torr\) So, the vapor pressure of the 0.378 m sodium diatrizoate solution at 31.2 °C is 33.9 torr.