Calculate the molality of a solution that contains 64.7 g of benzoic acid, C6H5COOH, in 427.1 mL of ethanol, C2H5OH. The density of ethanol is 0.789 g/mL.

Question

Chemistry

Charlotte Ayala

1 July, 10:32

Calculate the molality of a solution that contains 64.7 g of benzoic acid, C6H5COOH, in 427.1 mL of ethanol, C2H5OH. The density of ethanol is 0.789 g/mL.

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Answers (1)

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Adonis Proctor · Answer 1

The molality of the solution is 1.57 molal

Explanation:

Step 1: Data given

Mass of benzoic acid = 64.7 grams

Volume of ethanol = 427.1 mL

Density of ethanol = 0.789 g/mL

Molar mass benzoic acid = 122.12 g/mol

Step 2: Calculate moles benzoic acid

Moles benzoic acid = mass benzoic acid / molar mass benzoic acid

Moles benzoic acid = 64.7 grams / 122.12 g/mol

Moles benzoic acid = 0.530 moles

Step 3: Calculate mass ethanol

Mass ethanol = density * volume

Mass ethanol = 0.789 g/mL * 427.1 mL

Mass ethanol = 336.98 grams = 0.33698 kg

Step 4: Calculate molality

Molality = moles benzoic acid / mass ethanol

Molality = 0.530 moles / 0.33698 kg

Molality = 1.57 molal

The molality of the solution is 1.57 molal