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<p>Understanding Raoult's Law & Vapor Pressure Lowering</p>
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9. Solutions
Vapor Pressure Lowering (Raoult's Law)
Multiple Choice
How many grams of glucose, C6H12O6, must be added to 515.0 g of water to give a solution with a vapor pressure of 13.2 torr at 20.0ºC? The vapor pressure of pure water at 20.0ºC is 17.5 torr.
A
9.54 × 102 g
B
1.68 × 103 g
C
5.29 g
D
9.31 g
Verified step by step guidance1
Identify the given data: mass of solvent (water) = 515.0 g, vapor pressure of pure water (P°_solvent) = 17.5 torr, vapor pressure of solution (P_solution) = 13.2 torr, and temperature = 20.0ºC. The solute is glucose (C\_6H\_12O\_6).
Calculate the mole fraction of the solvent (water) using Raoult's Law: \(P_{solution} = X_{solvent} \times P^\circ_{solvent}\). Rearranged, this gives \(X_{solvent} = \frac{P_{solution}}{P^\circ_{solvent}}\).
Determine the moles of water (solvent) using its molar mass (approximately 18.015 g/mol): \(n_{water} = \frac{mass_{water}}{M_{water}}\).
Express the mole fraction of the solvent in terms of moles of solvent and solute: \(X_{solvent} = \frac{n_{water}}{n_{water} + n_{solute}}\). Use the value of \(X_{solvent}\) from step 2 and solve for \(n_{solute}\).
Calculate the mass of glucose (solute) by multiplying the moles of glucose by its molar mass (approximately 180.16 g/mol): \(mass_{solute} = n_{solute} \times M_{solute}\). This will give the grams of glucose needed.
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