1. A solution is prepared by mixing 1.0 grams of benzene (C6H6)in 100

1.	1. 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 molalityof benzene in the solution.
Answer» 1. 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

1. 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 molalityof benzene in the solution.

Answer»

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