The concentration of cholesterol dissolved in chloroform is \(6.15 \mathrm{~g}\) per \(100 \mathrm{ml}\) of solution. (a) A portion of this solution in a 5 -cm polarimeter tube causes an observed rotation of \(-1.2^{\circ}\). Calculate the specific rotation of cholesterol. (b) Predict the observed rotation if the same solution were placed in a \(10-\mathrm{cm}\) tube. (c) Predict the observed rotation if \(10 \mathrm{ml}\) of the solution were diluted to \(20 \mathrm{ml}\) and placed in a \(5-\mathrm{cm}\) tube.

First, let's find the concentration of cholesterol in the solution. We know that 6.15 g of cholesterol is dissolved in 100 ml of solution. To find the concentration in g/ml, we can divide 6.15 g by 100 ml: \(c = \frac{6.15 \mathrm {~g}}{100 \mathrm {~ml}} = 0.0615 \mathrm {~g/ml}\)

We are given that the observed rotation in the 5 -cm polarimeter tube is -1.2°.

Now we can use the formula for specific rotation to find the specific rotation of cholesterol: \[ [\alpha] = \frac{\alpha obs}{(l*c)}\] where \([\alpha]\) is the specific rotation, \(\alpha_{obs}\) is the observed rotation, \(l\) is the length of the polarimeter tube in decimeters (dm), and \(c\) is the concentration of the solution in g/ml. First, we need to convert the tube length from cm to dm: \(l = \frac {5 \mathrm {~cm}}{10 \mathrm {~cm/dm}} = 0.5 \mathrm {~dm}\) Now, we can calculate the specific rotation: \([\alpha] = \frac {-1.2^\circ}{0.5 \mathrm {~dm} * 0.0615 \mathrm {~g/ml}} = -39.022^\circ \mathrm {dm} / \mathrm {g} \cdot \mathrm {ml}\] Therefore, the specific rotation of cholesterol is -39.022° dm/g·ml.

For part (b), we need to find the observed rotation when the solution is placed in a 10-cm tube instead. We can rearrange the specific rotation formula to solve for the observed rotation: \(\alpha_{obs} = [\alpha] * (l*c)\) We need to convert the tube length from cm to dm: \(l = \frac {10 \mathrm {~cm}}{10 \mathrm {~cm/dm}} = 1 \mathrm {~dm}\) Now we can calculate the observed rotation: \(\alpha_{obs} = -39.022^\circ \mathrm {dm} / \mathrm {g} \cdot \mathrm {ml} * (1 \mathrm {~dm} * 0.0615 \mathrm {~g/ml}) = -2.4^\circ\) So, the observed rotation in the 10-cm tube is -2.4°.

For part (c), we need to find the observed rotation when 10 ml of the solution is diluted to 20 ml and placed in a 5-cm tube. First, we need to calculate the new concentration after dilution. Since the volume is doubled but the amount of cholesterol remains constant, the new concentration is half of the original concentration: \(c_{new} = 0.0615 \mathrm {~g/ml} * \frac {1}{2} = 0.03075 \mathrm {~g/ml}\) Now, we can use the formula for observed rotation with the new concentration and the 5-cm tube length (already converted in Step 3): \(\alpha_{obs, new} = -39.022^\circ \mathrm {dm} / \mathrm {g} \cdot \mathrm {ml} * (0.5 \mathrm {~dm} * 0.03075 \mathrm {~g/ml}) = -0.6^\circ\) So, the observed rotation for the diluted solution in a 5-cm tube is -0.6°.