Solutions and Colligative Properties Worksheet Molality, m is a measure of solve

Question

Solutions and Colligative Properties Worksheet Molality, m is a measure of solvent (kg) The mole frac ction of a solution is the same as that we learned for gases, and is defined as the amount in moles of the component in question divided by the total moles in the mixture. 1. Assume you dissolve 45.0 g of camphor, CoHieO, in 425 mL. of ethanol, CaH OH. Calculate the molality mole fraction, and weight ent of camphor in this solution. (The density of ethanol is 0.785 em.) 2. Fill in the blanks in the table. Aqueous solutions are assumed. Molarity D Molality ( Weight Pereent Mole Fraction 10.0 KNO CH-CO:H HOCH:CH:OH 0.0183 18.0 FeCb Total Ion Concentraion

Explanation / Answer

1. camphor = 45 g

moles camphor = 45 g/152.23 g/mol = 0.296 moles

mass ethanol = 425 ml = 0.425 kg

molality of solution = 0.296 moles/0.425 kg = 0.696 m

moles ethanol = 425 g/46.07 g/mol = 9.225 moles

Total moles = 0.296 + 9.225 = 9.521 moles

mole fraction camphor = 0.296/9.521 = 0.0311

Total mass = 425 + 45 = 470 g

wt% camphor = 45 x 100/470 = 9.57%

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2. Table

For 10 wt% KNO3 solution,

moles KNO3 = 100 g/101.1032 g/mol = 0.9891 moles

mass solution = 1000 g

mass wate = 900 g = 0.900 kg

moles water = 900 g/18 g/mol = 50 mmol

total moles = 50.9891 moles

molarity = 0.9891 moles/1 L = 0.9891 M

molality = 0.9891 moles/0.900 kg = 1.099 m

mole fraction = 0.9891/50.9891 = 0.0194

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For CH3CO2H,

molality = 0.0183 m

moles CH3CO2H = 0.0183 moles

mass CH3CO2H = 0.0183 moles x 60.05 g/mol = 1.1 g

mass water = 1000 - 1.1 = 998.9 g

molarity = 0.0183 moles/1.0 L = 0.0183 M

moles water = 989.9/18 = 55.5 moles

Total moles = 55.5 + 0.0183 = 55.5183 moles

mole fraction CH3CO2H = 0.0183/55.5183 = 0.00033

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For 18 wt% HOCH2CH2OH solution,

moles HOCH2CH2OH = 180 g/62.07 g/mol = 2.9 moles

mass water = 1000 - 180 = 820 g = 0.820 kg

moles water = 0.820 g/18 = 45.5 moles

molality of solution = 2.9 moles/0.820 kg = 3.54 m

molarity of solution = 2.9 moles/1 L = 2.9 M

Total moles = 48.4 moles

mole fraction HOCH2CH2OH = 2.9/48.4 = 0.06

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