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Adrenaline is the hormone that triggers the release of extra glucose molecules in times of stress or emergency. A solution of \(0.64 \mathrm{~g}\) of adrenaline in \(36.0 \mathrm{~g}\) of \(\mathrm{CCl}_{4}\) elevates the boiling point by \(0.49^{\circ} \mathrm{C}\). Calculate the approximate molar mass of adrenaline from this data.

Short Answer

Expert verified
\(Molar~mass~of~adrenaline = \frac{0.64~\text{g}}{\frac{0.49^{\circ}\mathrm{C} \cdot 0.036~\mathrm{kg}}{K_b}}\)

Step by step solution

01

Convert the mass of the solvent to moles

The molar mass of \(\mathrm{CCl}_4\) is 12.01 g/mol (C) + 4 * 35.45 g/mol (Cl) = 153.8 g/mol. Since we are given the mass of \(\mathrm{CCl}_4\) (36.0 g), we can convert this to moles: Moles of \(\mathrm{CCl}_4 = \frac{36.0 \,\text{g}}{153.8 \, \text{g/mol}} = 0.234 \, \mathrm{mol}\)
02

Use the formula for boiling point elevation

The formula for boiling point elevation is: \(\Delta T_b = K_b \cdot m\) Where \(\Delta T_b\) is the boiling point elevation, \(K_b\) is the molal boiling point elevation constant, and \(m\) is the molality of the solution. We're given that the boiling point elevation is \(0.49^{\circ}\mathrm{C}\), so we'll plug this value into the equation: \(0.49^{\circ}\mathrm{C} = K_b \cdot m\)
03

Calculate the molality of the solution

The molality (\(m\)) of a solution is defined as the number of moles of solute per kilogram of solvent. We have the number of moles of the solvent from Step 1, and we can convert the mass of the solvent to kilograms: Mass of solvent (\(\mathrm{CCl}_4\)) in kg = \(\frac{36.0 \,\text{g}}{1000 \, \text{g/kg}} = 0.036 \, \mathrm{kg}\) Now we can write the molality: \(m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}}\)
04

Solve for moles of solute (adrenaline)

Re-write the boiling point elevation formula from Step 2 with the molality expression: \(0.49^{\circ}\mathrm{C} = K_b \cdot \frac{\text{moles of solute}}{0.036 \, \mathrm{kg}}\) Now solve for the moles of solute (adrenaline), assuming that \(K_{b}\) is a known constant: \(\text{moles of solute} = \frac{0.49^{\circ}\text{C} \cdot 0.036 \,\text{kg}}{K_b}\)
05

Calculate the molar mass of adrenaline

We're given the mass of adrenaline (0.64 g) and now have the moles of adrenaline from Step 4. We can use this information to find the molar mass: Molar mass of adrenaline = \(\frac{\text{mass of adrenaline}}{\text{moles of adrenaline}}\) Plug in the values: Molar mass of adrenaline = \(\frac{0.64 \,\text{g}}{\frac{0.49^{\circ}\mathrm{C} \cdot 0.036 \, \mathrm{kg}}{K_b}}\) Assuming we have a value for \(K_{b}\), we can calculate the molar mass of adrenaline.

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Key Concepts

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

Boiling Point Elevation
When you dissolve a solute in a solvent, a curious phenomenon occurs: the boiling point of the solvent increases. This change is known as boiling point elevation. It can be quite handy in practical applications, such as calculating the molar mass of substances like adrenaline. The primary concept here centers around how certain properties of solutions aren't dependent on the chemical identity of the solute. Rather, they depend on the number of solute particles in the solution. To get a bit technical, boiling point elevation is calculated with the formula \(\Delta T_b = K_b \cdot m\), where \(\Delta T_b\) is the increase in boiling point, \(K_b\) is the ebullioscopic constant of the solvent, and \(m\) is the molality of the solution. The larger the number of particles, the greater the elevation in boiling point, helping us understand solution characteristics better.
Adrenaline
Adrenaline, also known as epinephrine, is a potent hormone responsible for the 'fight or flight' response in stressful situations. It stimulates the release of glucose into the bloodstream, providing a swift energy boost. In chemistry, adrenaline can be studied by examining its interactions in solutions, such as how it affects the boiling point. To discern more about adrenaline through experiments, one might dissolve it in a solvent such as carbon tetrachloride and observe how it alters the solvent's boiling conditions. By analyzing changes in physical properties, scientists can estimate important metrics like molar masses of soft molecules, such as adrenaline, aiding in understanding biochemical reactions and processes.
Colligative Properties
Colligative properties are key concepts in chemistry, relating to how properties of a solvent change once a solute is dissolved in it. What sets them apart is their reliance not on the solute’s type, but on its quantity. This includes properties like boiling point elevation we discussed earlier. The main colligative properties to know are:
  • Boiling point elevation
  • Freezing point depression
  • Vapor pressure lowering
  • Osmotic pressure
These properties help chemists determine the number of particles in a solution, which plays a crucial role in experiments and calculations, such as molar mass determinations of complex molecules. Understanding these properties allows for the practical prediction and manipulation of solution behavior essential for both academic and industrial applications.
CCl₄ (Carbon Tetrachloride)
Carbon tetrachloride, abbreviated as CCl₄, is an organic compound with historical importance in chemistry. It's a clear liquid widely used in the past as a cleaning agent and solvent. However, due to health and environmental concerns, its use has been limited. Despite this, CCl₄ still provides valuable insights when used in controlled laboratory environments to study solution dynamics. In the context of boiling point elevation, CCl₄ serves as a solvent to understand how solutes like adrenaline affect physical properties. Its well-known molar mass of 153.8 g/mol makes it a reliable benchmark. By examining how the boiling point changes, scientists can ascertain information about the solute, such as its molar mass—a critical task for understanding molecular properties and assisting in crafting safer chemical applications.

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Most popular questions from this chapter

Commercial concentrated aqueous ammonia is \(28 \% \mathrm{NH}_{3}\) by mass and has a density of \(0.90 \mathrm{~g} / \mathrm{mL}\). What is the molarity of this solution?

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