Predicting Colligative Properties for Electrolyte Solutions Predict the freezing point of aqueous 0.00145 m [latex]MgCl_2[/latex]. Analyze We will use a modified freezing-point depression equation in which the van't Hoff factor i is included. We first note that [latex]MgCl_2[/latex] is a salt that completely dissociates when it is dissolved in water. So we determine the value of i for [latex]MgCl_2[/latex]. We can do this by writing an equation to represent the dissociation of [latex]MgCl_2(s)[/latex]. Then we use the appropriate freezing-point depression expression.

[latex]MgCl _2( s ) \stackrel{ H _2 O }{\longrightarrow} Mg ^{2+}( aq ) + 2 Cl ^{-}( aq )[/latex] Because three moles of ions are obtained per mole of formula units dissolved, we expect the value i = 3. Now use the expression [latex]\Delta T_{ f }=-i \times K_{ f } \times m[/latex] [latex]= -3 \times 1.86 ^{\circ} C m ^{-1} \times 0.00145 m[/latex] = -0.0081 °C The predicted freezing point is -0.0081 °C . Assess Because the value of i for [latex]MgCl_2[/latex] is not exactly 3, we are not justified in carrying more than one or two significant figures in our answer. If we had ignored the fact that [latex]MgCl_2[/latex] is a strong electrolyte, then our calculated freezing-point depression would have been three times as small as the experimental value. Always remember to include the van't Hoff factor when ionic compounds are given as part of the problem.

Question 13.11: Predicting Colligative Properties for Electrolyte Solutions ...

General Chemistry: Principles and Modern Applications

Predicting Colligative Properties for Electrolyte Solutions

Predict the freezing point of aqueous 0.00145 m $MgCl_2$ .

Analyze
We will use a modified freezing-point depression equation in which the van’t Hoff factor i is included. We first note that $MgCl_2$ is a salt that completely dissociates when it is dissolved in water. So we determine the value of i for $MgCl_2$ . We can do this by writing an equation to represent the dissociation of $MgCl_2(s)$ . Then we use the appropriate freezing-point depression expression.

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$MgCl _2( s ) \stackrel{ H _2 O }{\longrightarrow} Mg ^{2+}( aq ) + 2 Cl ^{-}( aq )$

Because three moles of ions are obtained per mole of formula units dissolved, we expect the value i = 3. Now use the expression

$\Delta T_{ f }=-i \times K_{ f } \times m$

$= -3 \times 1.86 ^{\circ} C m ^{-1} \times 0.00145 m$

= -0.0081 °C

The predicted freezing point is -0.0081 °C.

Assess
Because the value of i for $MgCl_2$ is not exactly 3, we are not justified in carrying more than one or two significant figures in our answer. If we had ignored the fact that $MgCl_2$ is a strong electrolyte, then our calculated freezing-point depression would have been three times as small as the experimental value. Always remember to include the van’t Hoff factor when ionic compounds are given as part of the problem.