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

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
Transcript text: 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, C$_{27}$H$_{46}$O, 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 = CH$_3$CH$_2$OCH$_2$CH$_3$ = 74.12 g/mol. Mass = $\square$ g Submit Answer Retry Entire Group 9 more group attempts remaining
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Solution

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Solution Steps

Step 1: Determine the Change in Vapor Pressure

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: \[ \Delta P = 463.57 \, \text{mm Hg} - 454.07 \, \text{mm Hg} = 9.50 \, \text{mm Hg} \]

Step 2: Use Raoult's Law to Relate Vapor Pressure Change to Mole Fraction

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

Step 3: Calculate the Moles of Diethyl Ether

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: \[ n_1 = \frac{257.0 \, \text{g}}{74.12 \, \text{g/mol}} = 3.468 \, \text{mol} \]

Step 4: Relate Mole Fraction to Moles of Solute

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: \[ n_2 = X_2 \cdot n_1 = 0.0205 \cdot 3.468 = 0.0710 \, \text{mol} \]

Step 5: Convert Moles of Cholesterol to Grams

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

\[ \boxed{27.45 \, \text{g}} \]

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