Cholesterol (386.7 g/mol) is a nonelectrolyte and a critical component of the cell membranes of animal cells. What would be the change in boiling point of a solution containing 1.23 g of cholesterol and 105 g of chloroform (K_b = 3.63 $\circ$ C/m)? A)0.11 $\circ$ C B)0.22 $\circ$ C C)0.55 $\circ$ C D)1.1 $\circ$ C E)6.1 $\circ$ C

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To calculate the change in boiling point, use the formula ΔTb = i * Kb * m, where ΔTb is the change in boiling point, i is the van't Hoff factor (which is 1 for nonelectrolytes like cholesterol), Kb is the ebullioscopic constant of the solvent (chloroform, given as 3.63 °C/m), and m is the molality of the solution. First, find the moles of cholesterol: moles = 1.23 g / 386.7 g/mol = 0.00318 mol. Then, find the molality: molality = moles of solute / kg of solvent = 0.00318 mol / 0.105 kg = 0.0303 m. Finally, calculate ΔTb: ΔTb = 1 * 3.63 °C/m * 0.0303 m = 0.11 °C.

Cholesterol (386 ∘\circ∘ C/m)? A)0 ∘\circ∘ C B)0 ∘\circ∘ C C)0 ∘\circ∘ C D)1 ∘\circ∘

Cholesterol (386 $\circ$ C/m)?
A)0 $\circ$ C
B)0 $\circ$ C
C)0 $\circ$ C
D)1 $\circ$