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Multiple Choice
What is the vapor pressure of a 45.0% solution of glucose (C6H12O6) at a particular temperature, given that the vapor pressure of pure water at that temperature is 486 mm Hg?
A
268 mm Hg
B
267 mm Hg
C
267.5 mm Hg
D
267.3 mm Hg
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검증된 단계별 안내
1
Identify the solute and solvent in the solution. Here, glucose (C6H12O6) is the solute, and water is the solvent.
Use the formula for the vapor pressure of a solution: P_solution = X_solvent * P_pure_solvent, where P_solution is the vapor pressure of the solution, X_solvent is the mole fraction of the solvent, and P_pure_solvent is the vapor pressure of the pure solvent.
Calculate the mole fraction of the solvent (water). First, determine the mass of glucose and water in the solution. Since the solution is 45.0% glucose by mass, it means 45.0 g of glucose and 55.0 g of water in 100 g of solution.
Convert the masses of glucose and water to moles. Use the molar mass of glucose (C6H12O6) which is approximately 180.18 g/mol, and the molar mass of water (H2O) which is approximately 18.02 g/mol.
Calculate the mole fraction of water (X_water) using the formula: X_water = moles of water / (moles of water + moles of glucose). Substitute this value into the vapor pressure formula to find the vapor pressure of the solution.