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 $(K_a)$: $K_a = C \alpha^2 = 0.5 \times (0.075)^2 = 0.5 \times 0.005625 = 0.0028125 \approx 3 \times 10^{-3}$.

(B) The van't Hoff factor $i$ is defined as the ratio of theoretical molar mass to experimentally observed molar mass: $i = \frac{M_{\text{theoretical}}}{M_{\text{observed}}}$.
Given $i = 3$.
For $KCl \rightarrow K^+ + Cl^-$, $i = 2$.
For $K_2SO_4 \rightarrow 2K^+ + SO_4^{2-}$, $i = 3$.
For $MgSO_4 \rightarrow Mg^{2+} + SO_4^{2-}$, $i = 2$.
For $Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-}$, $i = 5$.
Thus, $K_2SO_4$ has $i = 3$.

(D) Step $1$: Calculate the depression in freezing point $(\Delta T_f)$.
$\Delta T_f = T_f^\circ - T_f = 0 \text{ }^\circ\text{C} - (-0.93 \text{ }^\circ\text{C}) = 0.93 \text{ K}$.
Step $2$: Use the formula for depression in freezing point with the van't Hoff factor:
$\Delta T_f = i \times K_f \times m$.
Step $3$: Rearrange the formula to solve for molality $(m)$:
$m = \frac{\Delta T_f}{i \times K_f}$.
Step $4$: Substitute the given values:
$m = \frac{0.93}{1.25 \times 1.86} = \frac{0.93}{2.325} = 0.4 \text{ m}$.
Thus, the molality of the solution is $0.4 \text{ m}$.

(C) The freezing point depression is given by $\Delta T_f = i \cdot K_f \cdot m$, where $i$ is the van't Hoff factor and $m$ is the molality.
For complete dissociation, $i$ equals the number of ions produced per formula unit.
$(a)$ $KCl \rightarrow K^+ + Cl^-$, $i = 2$. $\Delta T_f = 2 \times 0.2 = 0.4 \text{ m}$.
$(b)$ $NaCl \rightarrow Na^+ + Cl^-$, $i = 2$. $\Delta T_f = 2 \times 0.1 = 0.2 \text{ m}$.
$(c)$ $AlPO_4 \rightarrow Al^{3+} + PO_4^{3-}$, $i = 2$. $\Delta T_f = 2 \times 0.05 = 0.1 \text{ m}$.
$(d)$ $MgSO_4 \rightarrow Mg^{2+} + SO_4^{2-}$, $i = 2$. $\Delta T_f = 2 \times 0.15 = 0.3 \text{ m}$.
Comparing the values, $0.1 \text{ m}$ is the minimum value, which corresponds to $0.05 \text{ m Aluminium phosphate}$.

(B) The dissociation reaction for potassium ferrocyanide is:
$K_4[Fe(CN)_6] \rightarrow 4K^+ + [Fe(CN)_6]^{4-}$
Here, the number of ions produced per formula unit is $n = 5$.
The degree of dissociation $(\alpha)$ is $60\% = 0.6$.
The formula for the van't Hoff factor $(i)$ for dissociation is:
$i = 1 + \alpha(n - 1)$
Substituting the values:
$i = 1 + 0.6(5 - 1)$
$i = 1 + 0.6(4)$
$i = 1 + 2.4$
$i = 3.4$

(C) $1$. According to Raoult's law, adding a non-volatile solute decreases the vapour pressure of the solvent.
$2$. The decrease in vapour pressure leads to an elevation in the boiling point of the solvent.
$3$. Osmotic pressure is a colligative property, which depends on the number of particles. Electrolytes dissociate into multiple ions, increasing the number of particles compared to non-electrolytes at the same concentration, thus resulting in higher osmotic pressure.
$4$. Freezing point depression is a colligative property, but the freezing point itself is not.