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Chapter 11: Problem 71

Reserpine is a natural product isolated from the roots of the shrub Rauwolfia serpentina. It was first synthesized in 1956 by Nobel Prize winner \(\mathrm{R} . \mathrm{B}\). Woodward. It is used as a tranquilizer and sedative. When \(1.00 \mathrm{~g}\) reserpine is dissolved in \(25.0 \mathrm{~g}\) camphor, the freezing-point depression is \(2.63^{\circ} \mathrm{C}\left(K_{\mathrm{f}}\right.\) for camphor is \(40 .{ }^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol}\) ). Calculate the molality of the solution and the molar mass of reserpine.

Short Answer

Expert verified
The molality of the solution is \(0.06575 \, \text{mol/kg}\), and the molar mass of reserpine is approximately \(608.24 \, \text{g/mol}\).

Step by step solution

01

Understand the formula for freezing-point depression

The formula for freezing-point depression is given by: ΔTf = Kf * m where ΔTf is the freezing-point depression, Kf is the cryoscopic constant for the solvent (in this case, camphor), and m is the molality of the solution.
02

Calculate the molality of the solution using the given freezing-point depression and Kf value

We are given freezing-point depression (ΔTf) as 2.63°C and the Kf value of camphor as 40 °C·kg/mol. Now, we can rearrange the formula to solve for the molality (m): m = ΔTf / Kf m = 2.63 °C / 40 °C·kg/mol m = 0.06575 mol/kg
03

Use the molality of the solution to find the moles of solute (reserpine)

The molality (m) is defined as the moles of solute (reserpine) per kilogram of solvent (camphor). Therefore, we can use the molality to calculate the moles of reserpine based on the mass of camphor (25g): moles of reserpine = molality × mass of camphor / 1000 moles of reserpine = 0.06575 mol/kg × 25g / 1000 = 0.00164375 mol
04

Calculate the molar mass of reserpine

We are given the mass of reserpine as 1.00 g. Now, we can use the moles of reserpine calculated in the previous step to find the molar mass: Molar mass of reserpine = mass of reserpine / moles of reserpine Molar mass of reserpine = 1.00 g / 0.00164375 mol Molar mass of reserpine ≈ 608.24 g/mol So, the molality of the solution is 0.06575 mol/kg, and the molar mass of reserpine is approximately 608.24 g/mol.

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Most popular questions from this chapter

Is molality or molarity dependent on temperature? Explain your answer. Why is molality, and not molarity, used in the equations describing freezing-point depression and boilingpoint elevation?

A \(1.60-\mathrm{g}\) sample of a mixture of naphthalene \(\left(\mathrm{C}_{10} \mathrm{H}_{8}\right)\) and anthracene \(\left(\mathrm{C}_{14} \mathrm{H}_{10}\right)\) is dissolved in \(20.0 \mathrm{~g}\) benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{\mathrm{b}}\right)\). The freezing point of the solution is \(2.81^{\circ} \mathrm{C}\). What is the composition as mass percent of the sample mixture? The freezing point of benzene is \(5.51^{\circ} \mathrm{C}\) and \(K_{\mathrm{f}}\) is \(5.12^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol}\).

Which of the following statements is(are) true? Correct the false statements. a. The vapor pressure of a solution is directly related to the mole fraction of solute. b. When a solute is added to water, the water in solution has a lower vapor pressure than that of pure ice at \(0^{\circ} \mathrm{C}\). c. Colligative properties depend only on the identity of the solute and not on the number of solute particles present. d. When sugar is added to water, the boiling point of the solution increases above \(100^{\circ} \mathrm{C}\) because sugar has a higher boiling point than water.

The term proof is defined as twice the percent by volume of pure ethanol in solution. Thus, a solution that is \(95 \%\) (by volume) ethanol is 190 proof. What is the molarity of ethanol in a 92 proof ethanol-water solution? Assume the density ofethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\), is \(0.79 \mathrm{~g} / \mathrm{cm}^{3}\) and the density of water is \(1.0 \mathrm{~g} / \mathrm{cm}^{3}\)

Creatinine, \(\mathrm{C}_{4} \mathrm{H}_{7} \mathrm{~N}_{3} \mathrm{O}\), is a by-product of muscle metabolism, and creatinine levels in the body are known to be a fairly reliable indicator of kidney function. The normal level of creatinine in the blood for adults is approximately \(1.0 \mathrm{mg}\) per deciliter (dL) of blood. If the density of blood is \(1.025 \mathrm{~g} / \mathrm{mL}\), calculate the molality of a normal creatinine level in a \(10.0-\mathrm{mL}\) blood sample. What is the osmotic pressure of this solution at \(25.0^{\circ} \mathrm{C} ?\)
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