Questions: Use the References to access important values if needed for this question. The vapor pressure of diethyl ether (ether) is 463.57 mm Hg at 25°C. How many grams of cholesterol, C27H46O, a nonvolatile, nonelectrolyte (MW = 386.6 g/mol), must be added to 257.0 grams of diethyl ether to reduce the vapor pressure to 454.07 mm Hg? diethyl ether = CH3CH2OCH2CH3 = 74.12 g/mol. Mass = g

Solution Steps

The initial vapor pressure of diethyl ether is 463.57 mm Hg, and the final vapor pressure is 454.07 mm Hg. The change in vapor pressure (\(\Delta P\)) is: \[ \Delta P = 463.57 \, \text{mm Hg} - 454.07 \, \text{mm Hg} = 9.50 \, \text{mm Hg} \]

Raoult's Law for a nonvolatile solute states: \[ \Delta P = P_1^0 \cdot X_2 \] where \(P_1^0\) is the vapor pressure of the pure solvent (diethyl ether), and \(X_2\) is the mole fraction of the solute (cholesterol).

Rearranging for \(X_2\): \[ X_2 = \frac{\Delta P}{P_1^0} = \frac{9.50 \, \text{mm Hg}}{463.57 \, \text{mm Hg}} = 0.0205 \]

The mass of diethyl ether is 257.0 grams, and its molar mass is 74.12 g/mol. The number of moles of diethyl ether (\(n_1\)) is: \[ n_1 = \frac{257.0 \, \text{g}}{74.12 \, \text{g/mol}} = 3.468 \, \text{mol} \]

The mole fraction of the solute (cholesterol) is given by: \[ X_2 = \frac{n_2}{n_1 + n_2} \] where \(n_2\) is the number of moles of cholesterol. Since \(X_2\) is small, we can approximate: \[ X_2 \approx \frac{n_2}{n_1} \] Thus: \[ n_2 = X_2 \cdot n_1 = 0.0205 \cdot 3.468 = 0.0710 \, \text{mol} \]

The molar mass of cholesterol is 386.6 g/mol. The mass of cholesterol (\(m_2\)) is: \[ m_2 = n_2 \cdot \text{MW} = 0.0710 \, \text{mol} \cdot 386.6 \, \text{g/mol} = 27.45 \, \text{g} \]

Final Answer