## Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning

# Chapter 13 Solutions and Their Behavior - Study Questions - Page 505b: 37

#### Answer

See the answer below.

#### Work Step by Step

In 1 kg of the solution: Mass of solute: $3.00\%\times 1000\ g=30\ g$ Number of moles of solute: $30\ g\div 165.19\ g/mol=0.182\ mol$ Molality: $0.182\ mol\div 0.970\ kg=0.187\ m$ Assuming density is 1.0 kg/L $c=0.182\ mol/L$ a) Freezing point depression: $\Delta T_f=K_fm=-1.86^{\circ}C/m\times 0.187\ m=-0.35^{\circ}C$ Freezing point $0-0.35=-0.35^{\circ}C$ b) Boiling point elevation: $\Delta T_b=K_bm=0.5121^{\circ}C/m\times 0.187\ m=0.096^{\circ}C$ Boiling point $100+0.096=100.096^{\circ}C$ c) Osmotic pressure $\Pi=cRT=0.182\ mol/L\times 0.082\ L.atm/mol.K\times (25+273)K=4.44\ atm$ Since the osmotic pressure is very large, it's the one that can be measured the easiest with a small error.

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