16. (Ch6) At 25^circ C the vapor pressure of pure benzene (C_(6)H_(6)) is 0.1252 atm. How many moles of any nonvolatile, nondissoci ating solute must be added to 156 g of pure benzene at 25^circ C to result in a solution vapor pressure of 0.1000 atm ? a. 0.252 b. 0.337 c. 0.504 d. 0.674 e. 0.8

To solve this problem, we can use Raoult's Law, which states that the vapor pressure of a solvent in a solution is directly proportional to the mole fraction of the solvent in the solution. The formula for Raoult's Law is:<br /><br />\[ P_{\text{solution}} = X_{\text{solvent}} \cdot P_{\text{solvent}}^0 \]<br /><br />where:<br />- \( P_{\text{solution}} \) is the vapor pressure of the solution,<br />- \( X_{\text{solvent}} \) is the mole fraction of the solvent in the solution,<br />- \( P_{\text{solvent}}^0 \) is the vapor pressure of the pure solvent.<br /><br />Given:<br />- \( P_{\text{solvent}}^0 \) (vapor pressure of pure benzene) = 0.1252 atm<br />- \( P_{\text{solution}} \) (vapor pressure of the solution) = 0.1000 atm<br />- Mass of benzene = 156 g<br /><br />First, we need to find the number of moles of benzene. The molar mass of benzene (\( C_6H_6 \)) is approximately 78.11 g/mol.<br /><br />\[ \text{Moles of benzene} = \frac{\text{Mass of benzene}}{\text{Molar mass of benzene}} \]<br /><br />\[ \text{Moles of benzene} = \frac{156 \, \text{g}}{78.11 \, \text{g/mol}} \]<br /><br />\[ \text{Moles of benzene} \approx 2 \, \text{mol} \]<br /><br />Next, we use Raoult's Law to find the mole fraction of benzene in the solution:<br /><br />\[ X_{\text{benzene}} = \frac{P_{\text{solution}}}{P_{\text{solvent}}^0} \]<br /><br />\[ X_{\text{benzene}} = \frac{0.1000 \, \text{atm}}{0.1252 \, \text{atm}} \]<br /><br />\[ X_{\text{benzene}} \approx 0.796 \]<br /><br />Let \( n \) be the moles of the solute added. The total moles in the solution will be:<br /><br />\[ \text{Total moles} = \text{Moles of benzene} + n \]<br /><br />The mole fraction of benzene is then:<br /><br />\[ X_{\text{benzene}} = \frac{\text{Moles of benzene}}{\text{Total moles}} \]<br /><br />\[ 0.796 = \frac{2}{2 + n} \]<br /><br />Solving for \( n \):<br /><br />\[ 0.796(2 + n) = 2 \]<br /><br />\[ 1.592 + 0.796n = 2 \]<br /><br />\[ 0.796n = 2 - 1.592 \]<br /><br />\[ 0.796n = 0.408 \]<br /><br />\[ n \approx \frac{0.408}{0.796} \]<br /><br />\[ n \approx 0.513 \]<br /><br />Therefore, the number of moles of the solute needed is approximately 0.513 moles. The closest answer to this is:<br /><br />c. 0.504