For a 1.5 , m aqueous solution of Ca(NO_3)_2, | vedclass

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(B) The van't Hoff factor $i$ is given by the ratio of theoretical molar mass to observed molar mass: $i = \frac{M_{	ext{theoretical}}}{M_{	ext{obse

Vedclass · Accepted Answer

$0.75$

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(B) The van't Hoff factor $i$ is given by the ratio of theoretical molar mass to observed molar mass: $i = \frac{M_{	ext{theoretical}}}{M_{	ext{observed}}} = \frac{164}{65.4} \approx 2.5$. For the dissociation of $Ca(NO_3)_2 ightarrow Ca^{2+} + 2NO_3^-$,the number of ions produced per formula unit is $n = 3$. The relationship between $i$ and the degree of dissociation $\alpha$ is $i = 1 + (n - 1)\alpha$. Substituting the values: $2.5 = 1 + (3 - 1)\alpha$. $2.5 = 1 + 2\alpha$. $1.5 = 2\alpha$. $\alpha = \frac{1.5}{2} = 0.75$.

For a $1.5 \, m$ aqueous solution of $Ca(NO_3)_2$,the observed molar mass is $65.4 \, g \, mol^{-1}$ and the theoretical molar mass is $164 \, g \, mol^{-1}$. The degree of dissociation is .............

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