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Chapter 11: Problem 77

Reserpine is a natural product isolated from the roots of the shrub Rauwolfia serpentina. It was first synthesized in 1956 by Nobel Prize winner R. B. Woodward. It is used as a tranquilizer and sedative. When 1.00 g reserpine is dissolved in 25.0 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 reserpine and camphor solution is approximately \(0.06575\, \mathrm{mol/kg}\), and the molar mass of reserpine is approximately \(608.69\,\mathrm{g/mol}\).

Step by step solution

01

Write down the freezing-point depression formula

The freezing-point depression formula is given by: \[\Delta T_{f} = K_{f} \cdot m \cdot i\] where \(\Delta T_{f}\) is the freezing-point depression, \(K_{f}\) is the molal freezing-point depression constant (also known as the cryoscopic constant), \(m\) is the molality of the solution, and \(i\) is the van't Hoff factor (number of particles produced by a formula unit). In this case, since both reserpine and camphor are non-electrolytes, the van't Hoff factor is 1.
02

Calculate the molality of the solution

We are given that \(\Delta T_{f} = 2.63^{\circ}\mathrm{C}\), and the cryoscopic constant for camphor \(K_{f} = 40.0^{\circ}\mathrm{C}\cdot\mathrm{kg}/\mathrm{mol}\). Solving for the molality, we have: \[m = \frac{\Delta T_{f}}{K_{f} \cdot i} = \frac{2.63}{40.0 \cdot 1}\] Calculate the molality: \[m \approx 0.06575\, \mathrm{mol/kg}\]
03

Calculate the molar mass of reserpine

Now that we have the molality of the solution, we can use it to find the molar mass of reserpine. We have 1.00 g of reserpine dissolved in 25.0 g of camphor. First, convert the mass of camphor to kg: \[25.0\,\mathrm{g} \times \frac{1\,\mathrm{kg}}{1000\,\mathrm{g}} = 0.025\,\mathrm{kg}\] Now, using the molality, we can find the number of moles of reserpine in the solution: \[0.06575\,\mathrm{mol/kg} \times 0.025\,\mathrm{kg} \approx 0.00164375\,\mathrm{mol}\] Finally, we can find the molar mass of reserpine by dividing its mass by the number of moles: \[\frac{1.00\,\mathrm{g}}{0.00164375\,\mathrm{mol}} \approx 608.69\,\mathrm{g/mol}\] So, the molar mass of reserpine is approximately 608.69 g/mol.

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Most popular questions from this chapter

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