Question

QUESTION FOR BRAINLY EXPERT ‼️

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 following :-

Molarity
Mass percent
Mole fraction
Molality of Benzene in the solution

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Answer 1

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$\red{\bold{\underline{\underline{Molarity \:of \: Solution \::}}}}$

(1.0 grams benzene) / (78 g/mol) = 0.0128 moles benzene

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

$\red{\bold{\underline{\underline{Mass\: Percent \:of \: Solution \::}}}}$

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

$\red{\bold{\underline{\underline{Mole \: Fraction\:of \: Solution \::}}}}$

(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

$\red{\bold{\underline{\underline{Molality \:of \: Solution \::}}}}$

(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

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

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