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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 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?

Short Answer

Expert verified
The molar mass of sodium diatrizoate (NaDTZ) is 638 g/mol. The vapor pressure of the sodium diatrizoate solution at 31.2°C is 33.9 torr.

Step by step solution

01

Calculate the number of moles of NaDTZ

Given the mass of NaDTZ (38.4 g) and the molality (0.378 m), we can find the number of moles in the solution using the formula: Molality = moles of solute / kilogram of solvent Rearranging, we get: moles of solute = Molality * kilogram of solvent First, we need to convert the volume of water (160 mL) to be in grams: mass of water (g) = volume of water (mL) * density of water (g/mL) mass of water (g) = 1.60 * 10^2 mL * 0.995 g/mL = 159.2 g Next, convert the mass of water in grams to kilograms: mass of water (kg) = 159.2 g * (1 kg / 1000 g) = 0.1592 kg Now we can find the moles of solute: moles of NaDTZ = 0.378 m * 0.1592 kg = 0.0602 mol
02

Calculate the molar mass of NaDTZ

To calculate the molar mass of NaDTZ, we can use the formula: Molar mass = mass of solute / moles of solute Molar mass of NaDTZ = 38.4 g / 0.0602 mol = 638 g/mol
03

Calculate the mole fraction of NaDTZ and water in the solution

First, we need to find the moles of water in the solution. To do this, we can use the mass of water and the molar mass of water (18.015 g/mol): moles of water = mass of water / molar mass of water moles of water = 159.2 g / 18.015 g/mol = 8.84 mol Now, we can calculate the mole fraction of each component in the solution: Mole fraction of NaDTZ (X_NaDTZ) = moles of NaDTZ / (moles of NaDTZ + moles of water) X_NaDTZ = 0.0602 mol / (0.0602 mol + 8.84 mol) = 0.00678 Mole fraction of water (X_water) = moles of water / (moles of NaDTZ + moles of water) X_water = 8.84 mol / (0.0602 mol + 8.84 mol) = 0.993
04

Calculate the vapor pressure of the solution using Raoult's law

Raoult's law states that the vapor pressure of a component in a solution is proportional to its mole fraction: Vapor pressure of the solution = Vapor pressure of pure water * Mole fraction of water Vapor pressure of the solution = 34.1 torr * 0.993 = 33.9 torr The vapor pressure of the sodium diatrizoate solution at 31.2°C is 33.9 torr.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Molality
Molality is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per kilogram of solvent. Unlike molarity, which depends on the total volume of the solution, molality is only affected by the mass of the solvent. This makes it particularly useful when dealing with temperature changes since it does not change with volume fluctuations. To effectively calculate molality, you should first understand what moles are: a measure of the amount of a substance. Then, identify the solvent in the solution and ensure its mass is in kilograms for accurate results.
Demystifying Raoult's Law
Raoult's Law is essential when studying solutions and their vapor pressures. It states that the partial vapor pressure of each volatile component in a solution is directly proportional to its mole fraction. The total vapor pressure can be found by summing the products of the mole fraction of each component times its respective vapor pressure when pure. In simplest terms, Raoult's law helps us understand how each component contributes to the mixture's overall pressure, which is particularly important in predicting boiling points and studying colligative properties like boiling point elevation and freezing point depression.
Vapor Pressure in Solutions
Vapor pressure is a critical property of liquids. It represents the tendency of a substance to evaporate, influenced by temperature and the molecules' kinetic energy. Adding a solute to a solvent lowers the solvent's vapor pressure because the solute particles get in the way, making it more difficult for solvent molecules to escape into the vapor phase. This change in vapor pressure has notable effects on processes such as boiling and can help predict how solutions behave under various conditions. By calculating the mole fraction of the solute, as shown in the step by step solution, we can precisely determine the new vapor pressure of a solution using Raoult's law.
The Role of Mole Fraction
Mole fraction is a way of expressing the concentration of a component in a mixture. Calculated by dividing the number of moles of that component by the total number of moles of all substances present, it is a dimensionless quantity that plays a vital role in predicting properties like vapor pressure and boiling point of solutions. By knowing the mole fraction, scientists can also calculate partial pressures using Raoult's law, as seen in the solution provided. Understanding mole fraction can also simplify the comparison of different substances in a mixture, as it relates the amount of each substance rather than their mass or volume.

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Most popular questions from this chapter

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 \(\chi_{\mathrm{C}_{\mathrm{s}} \mathrm{H}_{\mathrm{l} 2} \mathrm{O}_{\mathrm{s}}}=0.01\) c. a solution of sodium chloride in water with \(\chi_{\mathrm{NaCl}}=0.01\) d. a solution of methanol in water with \(\chi_{\mathrm{CH_{3}}, \mathrm{OH}}=0.2\) (Consider the vapor pressure of both methanol \([143\) torr at \(\left.25^{\circ} \mathrm{C}\right]\) and water.

A solution is prepared by mixing 0.0300 mole of \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) and 0.0500 mole of \(\mathrm{CH}_{2} \mathrm{Br}_{2}\) at \(25^{\circ} \mathrm{C}\). Assuming the solution is ideal, calculate the composition of the vapor (in terms of molefractions) at \(25^{\circ} \mathrm{C} .\) At \(25^{\circ} \mathrm{C},\) the vapor pressures of pure \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) and pure \(\mathrm{CH}_{2} \mathrm{Br}_{2}\) are 133 and 11.4 torr, respectively.

Calculate the sodium ion concentration when \(70.0 \mathrm{mL}\) of 3.0 \(M\) sodium carbonate is added to \(30.0 \mathrm{mL}\) of \(1.0 \mathrm{M}\) sodium bicarbonate.

Plants that thrive in salt water must have internal solutions (inside the plant cells) that are isotonic with (have the same osmotic pressure as) the surrounding solution. A leaf of a saltwater plant is able to thrive in an aqueous salt solution (at \(25^{\circ} \mathrm{C}\) ) that has a freezing point equal to \(-0.621^{\circ} \mathrm{C}\). You would like to use this information to calculate the osmotic pressure of the solution in the cell. a. In order to use the freezing-point depression to calculate osmotic pressure, what assumption must you make (in addition to ideal behavior of the solutions, which we will assume)? b. Under what conditions is the assumption (in part a) reasonable? c. Solve for the osmotic pressure (at \(25^{\circ} \mathrm{C}\) ) of the solution in the plant cell. d. The plant leaf is placed in an aqueous salt solution (at \(\left.25^{\circ} \mathrm{C}\right)\) that has a boiling point of \(102.0^{\circ} \mathrm{C}\). What will happen to the plant cells in the leaf?

In a coffee-cup calorimeter, \(1.60 \mathrm{g} \mathrm{NH}_{4} \mathrm{NO}_{3}\) was mixed with \(75.0 \mathrm{g}\) water at an initial temperature \(25.00^{\circ} \mathrm{C}\). After dissolution of the salt, the final temperature of the calorimeter contents was \(23.34^{\circ} \mathrm{C}\) a. Assuming the solution has a heat capacity of \(4.18 \mathrm{J} / \mathrm{g} \cdot^{\circ} \mathrm{C}\) and assuming no heat loss to the calorimeter, calculate the enthalpy of solution \(\left(\Delta H_{\text {soln }}\right)\) for the dissolution of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) in units of kJ/mol. b. If the enthalpy of hydration for \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) is \(-630 . \mathrm{kJ} / \mathrm{mol}\), calculate the lattice energy of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\)

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