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Chapter 13: Problem 79

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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Most popular questions from this chapter

Which of the following in each pair is likely to be more soluble in hexane, \(\mathrm{C}_{6} \mathrm{H}_{14}:\) (a) \(\mathrm{CCl}_{4}\) or \(\mathrm{CaCl}_{2}\), (b) benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) or glycerol, $\mathrm{CH}_{2}(\mathrm{OH}) \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{OH},\( (c) octanoic acid, \)\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH},$ or acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH}\) ? Explain your answer in each case.

A solution contains \(0.50 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\) and an unknown number of moles of sodium chloride. The vapor pressure of the solution at \(29^{\circ} \mathrm{C}\) is \(3.85 \mathrm{kPa}\). The vapor pressure of pure water at this temperature is \(4.05 \mathrm{kPa}\). Calculate the number of grams of sodium chloride in the solution. (Hint: Remember that sodium chloride is a strong electrolyte.)

Proteins can be precipitated out of aqueous solution by the addition of an electrolyte; this process is called "salting out" the protein. (a) Do you think that all proteins would be precipitated out to the same extent by the same concentration of the same electrolyte? (b) If a protein has been salted out, are the protein-protein interactions stronger or weaker than they were before the electrolyte was added? (c) A friend of yours who is taking a biochemistry class says that salting out works because the waters of hydration that surround the protein prefer to surround the electrolyte as the electrolyte is added; therefore, the protein's hydration shell is stripped away, leading to protein precipitation. Another friend of yours in the same biochemistry class says that salting out works because the incoming ions adsorb tightly to the protein, making ion pairs on the protein surface, which end up giving the protein a zero net charge in water and therefore leading to precipitation. Discuss these two hypotheses. What kind of measurements would you need to make to distinguish between these two hypotheses?

Indicate whether each statement is true or false: (a) If you compare the solubility of a gas in water at two different temperatures, you find the gas is more soluble at the lower temperature. (b) The solubility of most ionic solids in water decreases as the temperature of the solution increases. (c) The solubility of most gases in water decreases as the temperature increases because water is breaking its hydrogen bonding to the gas molecules as the temperature is raised. (d) Some ionic solids become less soluble in water as the temperature is raised.

Calculate the number of moles of solute present in each of the following solutions: (a) \(255 \mathrm{~mL}\) of \(1.50 \mathrm{M} \mathrm{HNO}_{3}(a q)\) (b) \(50.0 \mathrm{mg}\) of an aqueous solution that is $1.50 \mathrm{~m} \mathrm{NaCl}$, (c) \(75.0 \mathrm{~g}\) of an aqueous solution that is \(1.50 \%\) sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) by mass.
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