29) A solution is prepared by dissolving 0.60 g of nicotine (a nonelectrolyte) in water to make 12 mL of solution. The osmotic pressure of the solution is 765.00kPa at 25^circ C The molecular weight of nicotine is g/mol A) 28 B) 43 C) 50 D) 160 E) 0.60 Answer: D

To solve this problem, we need to use the formula for osmotic pressure:<br /><br />\[ \pi = \frac{n}{V}RT \]<br /><br />Where:<br />- \(\pi\) is the osmotic pressure (in kPa)<br />- \(n\) is the number of moles of solute<br />- \(V\) is the volume of the solution (in liters)<br />- \(R\) is the ideal gas constant (in kPa·L/mol·K)<br />- \(T\) is the temperature (in Kelvin)<br /><br />Given:<br />- \(\pi = 765.00 \, \text{kPa}\)<br />- \(V = 12 \, \text{mL} = 0.012 \, \text{L}\)<br />- \(T = 25^{\circ}\text{C} = 298.15 \, \text{K}\)<br />- \(R = 8.314 \, \text{kPa·L/mol·K}\)<br /><br />First, we need to find the number of moles of nicotine (\(n\)) using the osmotic pressure formula:<br /><br />\[ n = \frac{\pi V}{RT} \]<br /><br />Substitute the given values into the equation:<br /><br />\[ n = \frac{765.00 \, \text{kPa} \times 0.012 \, \text{L}}{8.314 \, \text{kPa·L/mol·K} \times 298.15 \, \text{K}} \]<br /><br />\[ n = \frac{9.18 \, \text{kPa·L}}{2484.66 \, \text{kPa·L/mol}} \]<br /><br />\[ n \approx 0.00369 \, \text{mol} \]<br /><br />Next, we need to find the molecular weight of nicotine. The molecular weight (MW) is the mass of the solute divided by the number of moles:<br /><br />\[ \text{MW} = \frac{\text{mass}}{n} \]<br /><br />Given that the mass of nicotine is \(0.60 \, \text{g}\):<br /><br />\[ \text{MW} = \frac{0.60 \, \text{g}}{0.00369 \, \text{mol}} \]<br /><br />\[ \text{MW} \approx 162.38 \, \text{g/mol} \]<br /><br />Therefore, the molecular weight of nicotine is approximately \(160 \, \text{g/mol}\).<br /><br />The correct answer is:<br />D) 160