Colligative properties of electrolyte Questions in English

(A) The lowering of vapour pressure is a colligative property,which is directly proportional to the van't Hoff factor $(i)$ for the same molar concentration of solute.
$BaCl_2 \rightarrow Ba^{2+} + 2Cl^-$; $i = 3$
$NaCl \rightarrow Na^+ + Cl^-$; $i = 2$
$Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-}$; $i = 5$
Since the concentration is the same $(0.1 \ M)$ for all,the ratio of the lowering of vapour pressure is equal to the ratio of their van't Hoff factors.
Therefore,the ratio is $3 : 2 : 5$.

(A) The dissociation reaction for $BaCl_2$ is: $BaCl_{2(aq)} \rightarrow Ba^{2+}_{(aq)} + 2Cl^-_{(aq)}$
Initially,we have $1$ mole of $BaCl_2$.
Given the degree of dissociation $\alpha = 80\% = 0.8$.
At equilibrium,the moles are:
$BaCl_2 = 1 - \alpha = 1 - 0.8 = 0.2$
$Ba^{2+} = \alpha = 0.8$
$Cl^- = 2\alpha = 2 \times 0.8 = 1.6$
Total moles at equilibrium = $0.2 + 0.8 + 1.6 = 2.6$
The van't Hoff factor $i$ is defined as the ratio of total moles after dissociation to the initial moles:
$i = \frac{2.6}{1} = 2.6$

(B) The elevation in boiling point is a colligative property,which is directly proportional to the van't Hoff factor $(i)$ for solutions of the same normality $(N)$.
$KNO_3$ and $NaCl$ are strong electrolytes,so they dissociate completely into $2$ ions each $(i = 2)$.
$CH_3COOH$ is a weak electrolyte and dissociates partially,so its $i$ value is between $1$ and $2$ $(1 < i < 2)$.
Sucrose is a non-electrolyte,so it does not dissociate $(i = 1)$.
Since $i$ values are $KNO_3 = 2$,$NaCl = 2$,$CH_3COOH \approx 1.05$ (partial dissociation),and $\text{sucrose} = 1$,the order of boiling point is $1 \ N \ KNO_3 = 1 \ N \ NaCl > 1 \ N \ CH_3COOH > 1 \ N \ \text{sucrose}$.

(A) The chemical formula for Potassium trioxalatoaluminate $(III)$ is $K_3[Al(C_2O_4)_3]$.
When it dissociates completely in an aqueous solution,it breaks down as follows:
$K_3[Al(C_2O_4)_3] \rightarrow 3K^+ + [Al(C_2O_4)_3]^{3-}$.
The total number of ions produced from one formula unit of the compound is $3 + 1 = 4$.
Therefore,the Van't Hoff factor $(i)$,which represents the number of particles the solute splits into,is $4$.

(D) $1$. Calculate the number of moles of fluoroacetic acid: $\text{Moles} = \frac{19.5 \text{ g}}{78 \text{ g mol}^{-1}} = 0.25 \text{ mol}$.
$2$. Calculate the molality $(m)$: $m = \frac{0.25 \text{ mol}}{0.5 \text{ kg}} = 0.5 \text{ m}$.
$3$. Use the freezing point depression formula: $\Delta T_f = i \cdot K_f \cdot m$.
$4$. Given $\Delta T_f = 1 \text{ K}$,$K_f = 1.86 \text{ K kg mol}^{-1}$,and $m = 0.5 \text{ m}$,we have: $1 = i \times 1.86 \times 0.5$.
$5$. Solving for the van't Hoff factor $(i)$: $i = \frac{1}{0.93} \approx 1.075$.
$6$. Since $i = 1 + \alpha$ for a weak acid,$\alpha = i - 1 = 1.075 - 1 = 0.075$.
$7$. Calculate the dissociation constant $(K_a)$: $K_a = C \alpha^2 = 0.5 \times (0.075)^2 = 0.5 \times 0.005625 = 0.0028125 \approx 3 \times 10^{-3}$.