1. Calculate the molality of a solution that contains 56.5 g of benzoic acid, CH

Question

1. Calculate the molality of a solution that contains 56.5 g of benzoic acid, CHsCOOH, in 325 mL of ethanol, C-HsOH. The density of ethanol is 0.789 g/mL. 2. What are the mole fractions of ethanol, C2HsOH, and water in a solution prepared by mixing 75.0 ml. of ethanol with 25.0 mL of water at 25°C? The density of water is 1.00 g/ml. and the density of ethanol is 0.789 g/ml 3. Complete the following table for the aqueous solutions of potassium hydroxide. Density (g/ml.) 1.05 1.29 Mole Fraction of Solute Molarity Molality 1.13 0.121

Explanation / Answer

Solution :-

Molality = moles of solute/kg of solvent

Moles of benzoic acid = mass/molar mass

= 56.5g/122.12g/mol = 0.4627moles

Mass of ethanol = density*volume

= 0.789g/mL * 325 mL = 256.425g

Mass = 301.3g = 0.301kg

Molality = 0.4627moles/0.301kg = 1.54 m

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