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Question

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 iatrizoate, a nonvolatile water soluble compound. A 0.378-m solution is prepared by dissolving 38.4 g of sodium diatrizoate (NaDTZ) in 1.60×102 mL1.60 \times 10 ^ { 2 }\ \mathrm { mL } of water at 31.2C31.2 ^ { \circ } \mathrm { C } (the density of water at 31.2C31.2 ^ { \circ } \mathrm { C } is 0.995 g/mol). What is the molar mass of sodium diatrizoate? What is the vapor pressure of this solution if the vapor pressure of pure water at 31.2C31.2 ^ { \circ } \mathrm { C } is 34.1 torr?

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To determine the molar mass of sodium diatrizoate (NaDTZ) we need to know the mass and number of moles of sodium diatrizoate.

Before we calculate the moles of NaDTZ, we need to calculate the mass, and moles of water.

mass\textit{mass}(H2_2O) = d\textit{d} ×\times V\textit{V} = 0.995g/mL ×\times 160 mL = 0.159 kg

n\textit{n}(H2_2O) = massMolarmass\dfrac{mass}{Molar mass} = 159g18.016\dfrac{159g}{18.016} = 8.83 mol

We have the mass of NaDTZ - 38.4 g, so we need to determine the number of moles. The number of moles can be calculated from molality.

m\textit{m} = n(NaDTZ)mass(H2O\dfrac{n(NaDTZ)}{mass(H_2O}

n\textit{n}(NaDTZ) = m\textit{m} ×\times mass\textit{mass}(H2_2O)

n\textit{n}(NaDTZ) = 0.378 mol/kg ×\times 0.159 kg = 0.060 mol

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