The degree of dissociation of Ca(NO_3)_2 in a | vedclass

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(A) The molar mass of $Ca(NO_3)_2$ is $164 \ g/mol$. $7 \ g$ of $Ca(NO_3)_2 = \frac{7}{164} \approx 0.0427 \ mol$. $100 \ g$ of water $= \frac{100}{18

Vedclass · Accepted Answer

$746.1$

Vedclass · Answer

(A) The molar mass of $Ca(NO_3)_2$ is $164 \ g/mol$. $7 \ g$ of $Ca(NO_3)_2 = \frac{7}{164} \approx 0.0427 \ mol$. $100 \ g$ of water $= \frac{100}{18} \approx 5.56 \ mol$. Mole fraction of solute $(X) = \frac{0.0427}{0.0427 + 5.56} \approx 0.00762$. For $Ca(NO_3)_2 ightarrow Ca^{2+} + 2NO_3^-$,the number of ions $(n') = 3$. Degree of dissociation $(\alpha) = 0.7$. Van't Hoff factor $(i) = 1 + (n'-1)\alpha = 1 + (3-1) 	imes 0.7 = 1 + 1.4 = 2.4$. Using the formula for relative lowering of vapour pressure: $\frac{P^o - P}{P^o} = i 	imes X$. $\frac{760 - P}{760} = 2.4 	imes 0.00762 = 0.018288$. $760 - P = 760 	imes 0.018288 \approx 13.9$. $P = 760 - 13.9 = 746.1 \ mm \ Hg$.

The degree of dissociation of $Ca(NO_3)_2$ in a dilute aqueous solution containing $7 \ g$ of the salt per $100 \ g$ of water at $100^{\circ}C$ is $70$ percent. If the vapour pressure of water at $100^{\circ}C$ is $760 \ mm \ Hg$,calculate the vapour pressure of the solution.

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