## Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning

# Chapter 13 Solutions and Their Behavior - Study Questions - Page 505: 4

#### Answer

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#### Work Step by Step

KNO3 solution: For 1kg of the solution: Number of moles of solute: $100\ g\div101.10\ g/mol=0.989\ mol$ Molality: $0.989\ mol\div 0.9\ kg=1.099\ molal$ The number of moles of solvent: $900\ g\div 18.02\ g/mol=49.96\ mol$ Mole fraction: $0.989\div(0.989+49.96)=0.019$ Acetic acid solution: For 1kg of the solvent: Number of moles of solute: $0.0183\ molal\times1\ kg=0.0183\ mol$ Mass of solute: $0.0183\ mol\times 60.05\ g/mol=1.10\ g$ Weight percent: $1.10\div(1.10+1000)\times100\%=0.11\%$ Number of moles of solvent: $1000\ g\div 18.02\ g/mol=55.49\ mol$ Mole fraction: $0.0183\div(0.0183+55.49)=0.00033$ Ethylene glycol solution: For 1kg of the solution: Number of moles of solute: $180\ g\div62.07\ g/mol=2.90\ mol$ Molality: $2.90\ mol\div 0.82\ kg=3.54\ molal$ The number of moles of solvent: $820\ g\div 18.02\ g/mol=45.52\ mol$ Mole fraction: $2.90\div(2.90+45.52)=0.06$

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