Final answer:
The molarity of the benzene solution is 1.28 M, the mass percent is 9.09%, the mole fraction of benzene is 0.0225, and the molality is 1.28 m.
Step-by-step explanation:
The student is asking for the molarity, mass percent, mole fraction, and molality of benzene in a solution prepared by mixing 10 grams of benzene with 100 grams of water, resulting in a total volume of 100 mL. Unfortunately, the provided reference solution does not accurately apply to the student's question. However, the elements of the correct procedure are mentioned — such as calculating moles, utilizing solution volume for molarity, and modifying the molarity equation.
Given that benzene has a molar mass of approximately 78.11 g/mol, the number of moles of benzene in 10 grams can be calculated as follows:
Number of moles = Mass of benzene / Molar mass of benzene
Number of moles = 10 g / 78.11 g/mol = 0.128 moles of C6H6
To find the molarity (M), divide the moles of benzene by the total volume of the solution in liters:
M = moles of benzene / volume of solution in L
M = 0.128 mol / 0.1 L = 1.28 M
For mass percent, divide the mass of benzene by the total mass of the solution, then multiply by 100:
Mass percent = (mass of benzene / (mass of benzene + mass of water)) × 100%
Mass percent = (10 g / (10 g + 100 g)) × 100% = 9.09%
To calculate mole fraction for benzene, first calculate the moles of water:
Number of moles of water = Mass of water / Molar mass of water
Number of moles of water = 100 g / 18.015 g/mol = 5.55 moles of H2O
Mole fraction of benzene = moles of benzene / (moles of benzene + moles of water)
Mole fraction of benzene = 0.128 / (0.128 + 5.55) = 0.0225
For molality (b), divide the moles of benzene by the mass of the solvent (water) in kilograms:
Molality (b) = moles of benzene / mass of water in kg
Molality (b) = 0.128 mol / 0.1 kg = 1.28 m