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.
Answers
Answer:
hiii
your answer is here !
Explanation:
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
follow me !
Answer:
Explanation:
(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