Question

A solution is prepared by mixing 1.0 grams of benzene (C6H6) in 100 g of water to create a solution total volume of 100 ml. Calculate the molarity, mass percent, mole fraction, and molality of benzene in the solution.​

Answer 1

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Molarity:

(1.0 grams benzene) / (78 g/mol) = .0128 moles benzene

(0.128 moles0 / (0.100 L) = 1.28 Moles/Liter

Mass percent:

total mass = 1.0 g benzene + 100 g water = 101 g

1.0 grams benzene / (total mass)

1.0 g benezine / 101 g * 100 = .99 percent benzene

Mole fraction:

(1.0 grams benzene) / (78 g/mol) = .0128 moles benzene

(100 g water) / (16 g/mol) = 6.25 moles water

(.0128 moles benzene) / (.0128 moles benzene + 6.25 moles water) = .20 % benzene

Molality:

(1.0 grams benzene) / (78 g/mol) = .0128 moles benzene

(.0128 moles benzene) / (.1 kg water) = 1.28 mole/Kg

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Answer 2

Explanation:

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