Colligative properties of electrolyte Questions in English

(B) The dissociation of the weak acid is represented as: $HX \rightleftharpoons H^{+} + X^{-}$.
For the dissociation of an electrolyte,the van't Hoff factor $i$ is given by $i = 1 + \alpha(n-1)$,where $n$ is the number of ions produced per molecule.
Here,$n = 2$ and $\alpha = 0.20$ $(20 \%)$.
So,$i = 1 + 0.20(2-1) = 1.2$.
The lowering in freezing point is given by $\Delta T_{f} = i \times K_{f} \times m$.
Substituting the values: $\Delta T_{f} = 1.2 \times 1.86 \times 0.5 = 1.116 \ K \approx 1.12 \ K$.

(B) The osmotic pressure $(\pi)$ is a colligative property given by the formula $\pi = iCRT$,where $i$ is the van't Hoff factor,$C$ is the molar concentration,$R$ is the gas constant,and $T$ is the temperature.
For a given concentration and temperature,$\pi$ is directly proportional to the van't Hoff factor $(i)$.
$1$. Urea: $i = 1$ (non-electrolyte)
$2$. $CoCl_{2}$: $i = 3$ $(Co^{2+} + 2Cl^{-})$
$3$. $KCl$: $i = 2$ $(K^{+} + Cl^{-})$
$4$. Glucose: $i = 1$ (non-electrolyte)
Since $CoCl_{2}$ has the highest van't Hoff factor $(i = 3)$,it will exhibit the highest osmotic pressure.

(A) The dissociation of $K_{3}[Fe(CN)_{6}]$ is given by:
$K_{3}[Fe(CN)_{6}] \longrightarrow 3 K^{+} + [Fe(CN)_{6}]^{3-}$
Here,the number of ions produced per formula unit is $n = 3 + 1 = 4$.
The van't Hoff factor $(i)$ is calculated using the formula:
$i = 1 + \alpha(n - 1)$
Given the degree of dissociation $\alpha = 0.78$ and $n = 4$:
$i = 1 + 0.78(4 - 1) = 1 + 0.78(3) = 1 + 2.34 = 3.34$

(A) The depression in freezing point is given by the formula $\Delta T_{f} = i \times K_{f} \times m$.
Since $K_{f}$ and $m$ $(0.10 \ m)$ are constant for all solutions,$\Delta T_{f}$ depends directly on the van't Hoff factor $(i)$.
$i$ represents the number of particles produced upon dissociation.
For $Al_{2}(SO_{4})_{3}$,$i = 5$ $(2Al^{3+} + 3SO_{4}^{2-})$.
For $KI$,$i = 2$ $(K^{+} + I^{-})$.
For $C_{12}H_{22}O_{11}$ (sucrose) and $NH_{2}CONH_{2}$ (urea),$i = 1$ as they are non-electrolytes.
Since $Al_{2}(SO_{4})_{3}$ has the highest $i$ value,it will have the maximum $\Delta T_{f}$ value.

(B) Given: van't Hoff factor $i = 1.076$.
For the dissociation of mono-$1$-fluoroacetic acid: $CH_{2}FCOOH \rightleftharpoons CH_{2}FCOO^{-} + H^{+}$.
The number of ions produced per molecule is $n = 2$.
The formula for the degree of dissociation $\alpha$ is $\alpha = \frac{i-1}{n-1}$.
Substituting the values: $\alpha = \frac{1.076-1}{2-1} = \frac{0.076}{1} = 0.076$.

(D) The relationship between the van’t Hoff factor $(i)$ and the degree of dissociation $(\alpha)$ is given by:
$\alpha = \frac{i - 1}{n' - 1}$
Where,$n'$ is the number of ions formed after dissociation.
For the dissociation reaction: $A \rightleftharpoons n'B$
Initially: $1$ mole,$0$
After dissociation: $(1 - \alpha)$ mole,$n'\alpha$
Total number of moles present in the solution:
$= (1 - \alpha) + n'\alpha = 1 + (n' - 1)\alpha$
By definition,the van’t Hoff factor $(i)$ is the ratio of observed colligative property to calculated colligative property,which equals the ratio of total moles after dissociation to initial moles:
$i = \frac{1 + (n' - 1)\alpha}{1} = 1 + (n' - 1)\alpha$
Rearranging for $\alpha$:
$i - 1 = (n' - 1)\alpha$
$\alpha = \frac{i - 1}{n' - 1}$

(A) The dissociation of $Ba(NO_3)_2$ is given by: $Ba(NO_3)_2 \rightleftharpoons Ba^{2+} + 2NO_3^{-}$.
Here,the number of ions produced per formula unit is $n = 3$.
The formula for the degree of dissociation $(\alpha)$ in terms of the Van't Hoff factor $(i)$ is: $\alpha = \frac{i-1}{n-1}$.
Substituting the given values: $\alpha = \frac{2.74-1}{3-1} = \frac{1.74}{2} = 0.87$.

(D) The Van't Hoff factor $(i)$ represents the number of particles a solute dissociates into in a solution.
$Urea$ is a non-electrolyte,so it does not dissociate,and its $(i)$ value is $1$.
$AlCl_3$ dissociates as $AlCl_3 \rightarrow Al^{3+} + 3Cl^-$,so $(i) = 4$.
$K_2SO_4$ dissociates as $K_2SO_4 \rightarrow 2K^+ + SO_4^{2-}$,so $(i) = 3$.
$NH_4Cl$ dissociates as $NH_4Cl \rightarrow NH_4^+ + Cl^-$,so $(i) = 2$.
Therefore,$Urea$ has the lowest Van't Hoff factor.

(D) The elevation in boiling point is given by the formula $\Delta T_b = i \times K_b \times m$.
Since the concentration $(m)$ and the ebullioscopic constant $(K_b)$ are the same for all solutions,the boiling point depends on the van't Hoff factor $(i)$,which represents the number of particles produced upon dissociation.
$A$: Glucose $(C_6H_{12}O_6)$ is a non-electrolyte,so $i = 1$.
$B$: Sodium chloride $(NaCl)$ dissociates as $NaCl \rightarrow Na^+ + Cl^-$,so $i = 2$.
$C$: Calcium chloride $(CaCl_2)$ dissociates as $CaCl_2 \rightarrow Ca^{2+} + 2Cl^-$,so $i = 3$.
$D$: Ferric chloride $(FeCl_3)$ dissociates as $FeCl_3 \rightarrow Fe^{3+} + 3Cl^-$,so $i = 4$.
Since Ferric chloride has the highest value of $i$,it will have the highest boiling point.

(B) Depression in freezing point is a colligative property,which depends on the number of particles in the solution.
$KCl \rightleftharpoons K^{+} + Cl^{-}$ ($2$ particles)
$Na_{2}SO_{4} \rightleftharpoons 2Na^{+} + SO_{4}^{2-}$ ($3$ particles)
$MgSO_{4} \rightleftharpoons Mg^{2+} + SO_{4}^{2-}$ ($2$ particles)
$MgCO_{3} \rightleftharpoons Mg^{2+} + CO_{3}^{2-}$ ($2$ particles)
Since $Na_{2}SO_{4}$ produces the maximum number of particles ($3$ ions per formula unit),it causes the maximum depression in freezing point.

(A) The elevation in boiling point is a colligative property,which depends on the van't Hoff factor $(i)$,representing the number of particles produced upon dissociation.
For $0.1 \ M \ FeCl_3$,$i = 4$ $(Fe^{3+} + 3Cl^-)$.
For $0.1 \ M \ BaCl_2$,$i = 3$ $(Ba^{2+} + 2Cl^-)$.
For $0.1 \ M \ NaCl$,$i = 2$ $(Na^+ + Cl^-)$.
For $0.1 \ M \ \text{urea}$,$i = 1$ (non-electrolyte).
Since $0.1 \ M \ FeCl_3$ produces the maximum number of ions,it exhibits the highest elevation in boiling point and thus the highest boiling point.

(B) The van't Hoff factor $(i)$ for a strong electrolyte is equal to the total number of ions produced upon complete dissociation in an aqueous solution.
$1$. For $K_4[Fe(CN)_6]$: It dissociates as $K_4[Fe(CN)_6] \rightarrow 4K^+ + [Fe(CN)_6]^{4-}$. Total ions = $4 + 1 = 5$. Thus,$i = 5$.
$2$. For $Fe_4[Fe(CN)_6]_3$: It dissociates as $Fe_4[Fe(CN)_6]_3 \rightarrow 4Fe^{3+} + 3[Fe(CN)_6]^{4-}$. Total ions = $4 + 3 = 7$. Thus,$i = 7$.
$3$. For $[CoCl_2(en)_2]Cl$: It dissociates as $[CoCl_2(en)_2]Cl \rightarrow [CoCl_2(en)_2]^+ + Cl^-$. Total ions = $1 + 1 = 2$. Thus,$i = 2$.
Therefore,the values are $5, 7, 2$.

(C) The boiling point elevation is a colligative property,which depends on the van't Hoff factor $(i)$ of the solute. The formula is $\Delta T_b = i \times K_b \times m$.
Since the molality $(m)$ is the same for all solutions,the solution with the highest van't Hoff factor $(i)$ will have the highest boiling point.
$1.$ $KNO_3 \rightarrow K^+ + NO_3^-$,$i = 2$.
$2.$ $\text{Urea}$ is a non-electrolyte,$i = 1$.
$3.$ $K_4[Fe(CN)_6] \rightarrow 4K^+ + [Fe(CN)_6]^{4-}$,$i = 5$.
$4.$ $NH_4NO_3 \rightarrow NH_4^+ + NO_3^-$,$i = 2$.
Comparing the values,$K_4[Fe(CN)_6]$ has the highest $i$ value $(i = 5)$,therefore it has the highest boiling point.

(C) The Van't Hoff factor $(i)$ for a strong electrolyte like $NaCl$ is given by the formula $i = 1 + (n-1)\alpha$,where $n$ is the number of ions produced per formula unit and $\alpha$ is the degree of dissociation.
For $NaCl$,$n = 2$. As the concentration of the solution decreases,the degree of dissociation $(\alpha)$ increases due to the decrease in inter-ionic attractions.
Therefore,as concentration decreases from $0.1 \ M$ to $0.001 \ M$,$\alpha$ increases,which leads to an increase in the value of $i$.
Thus,the order of the Van't Hoff factor is $i_{A} < i_{B} < i_{C}$.

(A) The dissociation of $CaCl_2$ is represented as: $CaCl_2 \rightarrow Ca^{2+} + 2Cl^-$.
Initially,let the number of moles be $1$.
After $80\%$ dissociation,the moles are:
$CaCl_2 = 1 - 0.8 = 0.2$
$Ca^{2+} = 0.8$
$Cl^- = 2 \times 0.8 = 1.6$
Total moles after dissociation = $0.2 + 0.8 + 1.6 = 2.6$.
The Van't Hoff factor $(i)$ is defined as the ratio of total moles after dissociation to the initial moles.
$i = \frac{2.6}{1} = 2.6$.
Therefore,the correct option is $A$.

(B) The dissociation of $K_3[Fe(CN)_6]$ is given by:
$K_3[Fe(CN)_6] \rightarrow 3K^+ + [Fe(CN)_6]^{3-}$
Here,the number of ions produced per formula unit $(n)$ is $3 + 1 = 4$.
The relationship between the Van't Hoff factor $(i)$,degree of dissociation $(\alpha)$,and the number of ions $(n)$ is:
$i = 1 + \alpha(n - 1)$
Given $\alpha = 0.778$ and $n = 4$:
$i = 1 + 0.778(4 - 1)$
$i = 1 + 0.778(3)$
$i = 1 + 2.334$
$i = 3.334$
Therefore,the correct option is $B$.

(B) The formula for depression in freezing point is $\Delta T_f = i \times K_f \times m$.
Given: $\Delta T_f = 0.0744 \ K$,$K_f = 1.86 \ K \ kg \ mol^{-1}$,and $m = 0.01 \ m$.
Substituting the values: $0.0744 = i \times 1.86 \times 0.01$.
$i = \frac{0.0744}{0.0186} = 4$.
For complete dissociation of $K_x[Fe(CN)_6]$,the van't Hoff factor $i$ is equal to the number of ions produced,which is $x + 1$.
$x + 1 = 4 \Rightarrow x = 3$.
Therefore,the molecular formula is $K_3[Fe(CN)_6]$.

(A) The elevation in boiling point is a colligative property,which depends on the van't Hoff factor $(i)$.
$\Delta T_b = i \times K_b \times m$.
For a given molality $(m)$ and solvent,$\Delta T_b$ is directly proportional to $i$.
- For $1 \ m \ FeCl_3$,$i = 4$ $(Fe^{3+} + 3Cl^-)$.
- For $1 \ m \ NaCl$,$i = 2$ $(Na^+ + Cl^-)$.
- For $1 \ m \ BaCl_2$,$i = 3$ $(Ba^{2+} + 2Cl^-)$.
- For $1 \ m \ \text{urea}$,$i = 1$ (non-electrolyte).
Since $FeCl_3$ has the highest van't Hoff factor $(i = 4)$,it will show the highest elevation in boiling point,and thus the highest boiling point. The correct option is $A$.

(C) The elevation in boiling point is given by the formula $\Delta T_b = i \times K_b \times m$.
For $0.01 \ M$ urea (a non-electrolyte),the van't Hoff factor $i = 1$.
For $0.01 \ M \ BaCl_2$ (a strong electrolyte),it dissociates as $BaCl_2 \rightarrow Ba^{2+} + 2Cl^-$,so the van't Hoff factor $i = 3$.
Since the molality $m$ and the ebullioscopic constant $K_b$ are the same for both solutions,the ratio of elevation in boiling point is $\frac{\Delta T_b(BaCl_2)}{\Delta T_b(\text{urea})} = \frac{i(BaCl_2)}{i(\text{urea})} = \frac{3}{1} = 3$.
Therefore,the elevation in boiling point of $0.01 \ M \ BaCl_2$ is approximately three times that of $0.01 \ M$ urea.

(B) The boiling point elevation is given by $\Delta T_b = i \times K_b \times m$. The solution with the highest boiling point will have the highest value of the van't Hoff factor $(i)$ multiplied by molality $(m)$.
For $A$: $i = 2$ $(NaCl \rightarrow Na^+ + Cl^-)$,$m = 0.1$,so $i \times m = 0.2$.
For $B$: $i = 3$ $(Ba(NO_3)_2 \rightarrow Ba^{2+} + 2NO_3^-)$,$m = 0.2$,so $i \times m = 0.6$.
For $C$: $i = 4$ $(Na_3PO_4 \rightarrow 3Na^+ + PO_4^{3-})$,$m = 0.01$,so $i \times m = 0.04$.
For $D$: $i = 2$ $(KNO_3 \rightarrow K^+ + NO_3^-)$,$m = 0.03$,so $i \times m = 0.06$.
Comparing the values,$0.6$ is the highest,therefore $0.2 \ m$ $Ba(NO_3)_2$ has the highest boiling point.

(A) The boiling point elevation is given by the formula $\Delta T_b = i \times K_b \times m$.
Since the molality $(m)$ and the ebullioscopic constant $(K_b)$ are identical for all solutions,the boiling point depends on the van't Hoff factor $(i)$.
For $K_4[Fe(CN)_6]$,$i = 5$ $(4K^+ + [Fe(CN)_6]^{4-})$.
For $Na_2SO_4$,$i = 3$ $(2Na^+ + SO_4^{2-})$.
For Urea,$i = 1$ (non-electrolyte).
For $NaCl$,$i = 2$ $(Na^+ + Cl^-)$.
Since $K_4[Fe(CN)_6]$ has the highest van't Hoff factor $(i=5)$,it will show the highest boiling point elevation and thus the highest boiling point.

(B) The boiling point elevation is given by the formula $\Delta T_b = i \times K_b \times m$. Since $K_b$ and $m$ are constant for the given solutions,the boiling point elevation depends on the van't Hoff factor $(i)$.
For $KNO_3$,it dissociates as $K^+ + NO_3^-$,so $i = 2$.
For $0.1 \ m$ Urea,$i = 1$ (non-electrolyte).
For $0.1 \ m$ sodium chloride $(NaCl)$,it dissociates as $Na^+ + Cl^-$,so $i = 2$.
For $0.1 \ m$ potassium sulfate $(K_2SO_4)$,it dissociates as $2K^+ + SO_4^{2-}$,so $i = 3$.
For $0.1 \ m$ aluminum nitrate $(Al(NO_3)_3)$,it dissociates as $Al^{3+} + 3NO_3^-$,so $i = 4$.
Therefore,the boiling point elevation of $0.1 \ m$ $KNO_3$ is equal to that of $0.1 \ m$ sodium chloride.

(C) For a solute dissociating as $A \rightarrow nB$,the initial moles are $1$ and $0$ respectively.
After dissociation,the moles are $(1-\alpha)$ and $n\alpha$.
The total number of moles at equilibrium is $(1-\alpha + n\alpha) = 1 + \alpha(n-1)$.
The van't Hoff factor $i$ is defined as the ratio of observed moles to initial moles: $i = \frac{1 + \alpha(n-1)}{1}$.
Rearranging for $\alpha$: $i - 1 = \alpha(n-1)$.
Therefore,$\alpha = \frac{i-1}{n-1}$.

(B) The elevation in boiling point is a colligative property,which depends on the total number of solute particles in the solution.
For solutions with the same molar concentration,the one that produces the highest number of ions upon dissociation will exhibit the highest boiling point.

Aqueous Solution	Number of Ions Produced
$1.0 \ M \ NaOH \rightarrow Na^{+} + OH^{-}$	$2$
$1.0 \ M \ Na_{2}SO_{4} \rightarrow 2Na^{+} + SO_{4}^{2-}$	$3$
$1.0 \ M \ NH_{4}NO_{3} \rightarrow NH_{4}^{+} + NO_{3}^{-}$	$2$
$1.0 \ M \ KNO_{3} \rightarrow K^{+} + NO_{3}^{-}$	$2$

Since $Na_{2}SO_{4}$ produces the maximum number of ions ($3$ ions per formula unit),it will have the highest boiling point.

(D) The boiling point elevation is given by the formula $\Delta T_{b} = i \times K_{b} \times m$. Since the molality $(m)$ and the ebullioscopic constant $(K_{b})$ are the same for all solutions,$\Delta T_{b} \propto i$,where $i$ is the van't Hoff factor.
For complete ionization:
$(A)$ $AlCl_3 \rightarrow Al^{3+} + 3Cl^-$,so $i = 4$.
$(B)$ $Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-}$,so $i = 5$.
$(C)$ $K_2SO_4 \rightarrow 2K^+ + SO_4^{2-}$,so $i = 3$.
$(D)$ $NaCl \rightarrow Na^+ + Cl^-$,so $i = 2$.
The solution with the lowest $i$ value will exhibit the lowest boiling point elevation. Thus,$NaCl$ has the lowest boiling point elevation.

(A) $1$. Calculate molality $(m)$ of each solute in $100 \ g$ of solution (assuming $84.65 \ g$ of water solvent):
$m_{AB} = \frac{5.85 \ g / 58.5 \ g \ mol^{-1}}{0.08465 \ kg} = 1.181 \ mol \ kg^{-1}$
$m_{XY_2} = \frac{9.50 \ g / 95 \ g \ mol^{-1}}{0.08465 \ kg} = 1.181 \ mol \ kg^{-1}$
$2$. Calculate van't Hoff factor $(i)$ for each:
For $AB \rightarrow A^+ + B^-$, $i_1 = 1 + (2-1)0.8 = 1.8$
For $XY_2 \rightarrow X^{2+} + 2Y^-$, $i_2 = 1 + (3-1)0.6 = 2.2$
$3$. Calculate total depression in freezing point $(\Delta T_f)$:
$\Delta T_f = K_f \times (i_1 m_1 + i_2 m_2) = 1.86 \times (1.8 \times 1.181 + 2.2 \times 1.181) = 1.86 \times (4.724) \approx 8.786 \ K$
$4$. Calculate final freezing point:
$T_f = 273 \ K - 8.786 \ K = 264.214 \ K \approx 264.25 \ K$.

(A) When an electrolyte undergoes dissociation in a solution,it splits into multiple ions or particles.
This increase in the total number of solute particles in the solution leads to an increase in the observed colligative properties compared to the calculated values.
Since the van't Hoff factor $(i)$ is defined as the ratio of observed colligative property to the calculated colligative property,for dissociation,$i > 1$.

(D) The depression in freezing point is given by the formula $\Delta T_f = i \times K_f \times m$.
Since the molality $(m)$ and the cryoscopic constant $(K_f)$ are the same for all solutions,the depression in freezing point $(\Delta T_f)$ is directly proportional to the van't Hoff factor $(i)$.
The freezing point of the solution is $T_f = T_f^0 - \Delta T_f$. Therefore,the solution with the lowest value of $i$ will have the highest freezing point.

Compound	van't Hoff Factor $(i)$
$Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-}$	$5$
$BaCl_2 \rightarrow Ba^{2+} + 2Cl^{-}$	$3$
$AlCl_3 \rightarrow Al^{3+} + 3Cl^{-}$	$4$
$NH_4Cl \rightarrow NH_4^+ + Cl^{-}$	$2$

Comparing the values of $i$,$NH_4Cl$ has the lowest van't Hoff factor $(i = 2)$. Thus,it will have the minimum depression in freezing point and consequently the highest freezing point.

(A) For a centimolar solution,the concentration $C = 0.01 \ M = 10^{-2} \ mol \ L^{-1}$.
Acetic acid $(CH_3COOH)$ dissociates as: $CH_3COOH \rightleftharpoons CH_3COO^- + H^+$.
The degree of dissociation $\alpha = 0.5$.
The van't Hoff factor $i = 1 + \alpha(n-1)$,where $n=2$.
$i = 1 + 0.5(2-1) = 1.5$.
The osmotic pressure $\pi = iCRT$.
Given $T = 27 + 273 = 300 \ K$ and $R = 0.083 \ L \ atm \ K^{-1} \ mol^{-1}$.
$\pi = 1.5 \times 0.01 \times 0.083 \times 300$.
$\pi = 1.5 \times 0.01 \times 24.9 = 1.5 \times 0.249 = 0.3735 \ atm$.
Rounding to the nearest option,the value is $0.37 \ atm$.

(C) The formula for the depression in freezing point is given by $\Delta T_{f} = i \times K_{f} \times m$.
Given:
van't Hoff factor $(i) = 1.075$
Molality $(m) = 0.5 \ m$
Freezing point depression constant $(K_{f}) = 1.86 \ K \ kg \ mol^{-1}$
Substituting the values into the formula:
$\Delta T_{f} = 1.075 \times 1.86 \times 0.5$
$\Delta T_{f} = 1.075 \times 0.93$
$\Delta T_{f} = 0.99975 \ K \approx 1.0 \ K$.

(A) Given: $\Delta T_{b} = 0.01 \ K$,molality $(m) = 0.01 \ m$,and $K_{b} = 0.52 \ K \ kg \ mol^{-1}$.
We use the formula for elevation in boiling point: $\Delta T_{b} = i \times K_{b} \times m$.
Rearranging to solve for the Van't Hoff factor $(i)$: $i = \frac{\Delta T_{b}}{K_{b} \times m}$.
Substituting the values: $i = \frac{0.01}{0.52 \times 0.01} = \frac{1}{0.52} \approx 1.92$.
Thus,the Van't Hoff factor is $1.92$.

(C) The chemical formula for Ferric sulphate is $Fe_2(SO_4)_3$.
Upon $100 \%$ ionization,it dissociates as follows:
$Fe_2(SO_4)_3 \rightarrow 2Fe^{3+} + 3SO_4^{2-}$
The van't Hoff factor $(i)$ is the total number of ions produced per formula unit.
$i = 2 + 3 = 5$.

(D) The depression in freezing point is given by $\Delta T_{f} = T_{f}^{\circ} - T_{f}$.
For glucose (a non-electrolyte),$\Delta T_{f} = 0^{\circ} C - (-0.0186^{\circ} C) = 0.0186^{\circ} C$.
Since $\Delta T_{f} = i \times m \times K_{f}$ and both solutions have the same molarity $(10^{-3} \ M)$,their molality $(m)$ is also the same.
For $NaCl$,the van't Hoff factor $i = 2$ (as $NaCl \rightarrow Na^{+} + Cl^{-}$).
Therefore,$\Delta T_{f}(NaCl) = i \times \Delta T_{f}(\text{glucose}) = 2 \times 0.0186^{\circ} C = 0.0372^{\circ} C$.
The freezing point of the $NaCl$ solution is $T_{f} = T_{f}^{\circ} - \Delta T_{f} = 0^{\circ} C - 0.0372^{\circ} C = -0.0372^{\circ} C$.

(A) Given: Weight of solute $= 10 \ g$,Weight of solvent $= 100 \ g$,Molecular mass of solute $= 100 \ g/mol$.
Since it is a binary electrolyte,the van't Hoff factor $i = 2$.
The formula for elevation in boiling point is $\Delta T_b = i \times K_b \times m$,where $m$ is the molality.
$m = \frac{\text{weight of solute}}{\text{molecular mass}} \times \frac{1000}{\text{weight of solvent in g}} = \frac{10}{100} \times \frac{1000}{100} = 1 \ m$.
Substituting the values: $\Delta T_b = 2 \times K_b \times 1$.
Therefore,$K_b = \frac{\Delta T_b}{2}$.

(C) The depression in freezing point is a colligative property,which depends on the number of particles in the solution.
Glucose is a non-electrolyte and does not dissociate in water,so its van't Hoff factor $(i)$ is $1$.
$MgCl_2$ is a strong electrolyte that dissociates as $MgCl_2 \rightarrow Mg^{2+} + 2Cl^-$,producing $3$ ions per formula unit,so its van't Hoff factor $(i)$ is $3$.
Since the concentration of both solutions is $0.01 \ M$,the depression in freezing point for $MgCl_2$ will be approximately $3$ times that of the glucose solution.
Therefore,the correct option is $(C)$.

(D) The freezing point depression is a colligative property,which depends on the van't Hoff factor $(i)$,representing the number of particles produced per formula unit in solution.
Urea is a non-electrolyte,so $i = 1$.
$NaCl$ dissociates as $NaCl \rightarrow Na^+ + Cl^-$,so $i = 2$.
$Na_2SO_4$ dissociates as $Na_2SO_4 \rightarrow 2Na^+ + SO_4^{2-}$,so $i = 3$.
$Al_2(SO_4)_3$ dissociates as $Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-}$,so $i = 5$.
Since the freezing point depression $\Delta T_f = i \times K_f \times m$,the solution with the highest value of $i$ will have the maximum depression in freezing point,resulting in the lowest freezing point.
Thus,$1 \ M \ Al_2(SO_4)_3$ has the lowest freezing point. Hence,the correct option is $D$.

(C) The elevation in boiling point is given by the formula: $\Delta T_{b} = i \times K_{b} \times m$.
Given: $\Delta T_{b} = 0.01 \ K$,$i = 1.92$,$K_{b} = 0.52 \ K \ kg \ mol^{-1}$.
Rearranging the formula for molality $(m)$: $m = \frac{\Delta T_{b}}{i \times K_{b}}$.
Substituting the values: $m = \frac{0.01}{1.92 \times 0.52} = \frac{0.01}{0.9984} \approx 0.010 \ m$.

(B) Given: Osmotic pressure $\pi = 3.735 \times 10^{-3} \ bar$,mass of $K_2SO_4$ $(\omega) = 17.4 \ mg = 17.4 \times 10^{-3} \ g$,volume $(V) = 2 \ L$,temperature $(T) = 27^{\circ} C = 300 \ K$.
Molar mass $(M)$ of $K_2SO_4 = (2 \times 39) + 32 + (4 \times 16) = 174 \ g \ mol^{-1}$.
Using the formula $\pi = iCRT$,where $C = \frac{\omega}{M \times V}$,we get $i = \frac{\pi \times M \times V}{\omega \times R \times T}$.
Substituting the values: $i = \frac{3.735 \times 10^{-3} \times 174 \times 2}{17.4 \times 10^{-3} \times 0.083 \times 300} = \frac{1.30038}{0.43326} = 3.0$.
Thus,the van't Hoff factor is $3$.

(B) The elevation in boiling point $(\Delta T_b)$ is given by the formula: $\Delta T_b = i \times K_b \times m$.
Since the molality $(m)$ and the ebullioscopic constant $(K_b)$ are the same for all solutions,$\Delta T_b$ is directly proportional to the van't Hoff factor $(i)$.
For $Na_2SO_4$ (sodium sulphate),$i = 3$ (as it dissociates into $2Na^+ + SO_4^{2-}$).
For urea (a non-electrolyte),$i = 1$.
For $NaCl$ (sodium chloride),$i = 2$ (as it dissociates into $Na^+ + Cl^-$).
Therefore,the ratio of the elevation in boiling point is $3 : 1 : 2$.

(A) Step $I$: Calculation of molality $(m)$
Molar mass of $CH_3CH_2CHClCOOH = 122.5 \ g \ mol^{-1}$.
Molality $(m)$ = $\frac{\text{Mass of solute}}{\text{Molar mass of solute} \times \text{Mass of solvent in kg}} = \frac{12.25 \ g}{122.5 \ g \ mol^{-1} \times 0.250 \ kg} = 0.40 \ m$.
Step $II$: Calculation of degree of dissociation $(\alpha)$
For a weak acid,$K_a = C\alpha^2$ (assuming $\alpha \ll 1$).
$\alpha = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{1.44 \times 10^{-3}}{0.4}} = \sqrt{36 \times 10^{-4}} = 0.06$.
Step $III$: Calculation of van't Hoff factor $(i)$
For dissociation $CH_3CH_2CHClCOOH \rightleftharpoons CH_3CH_2CHClCOO^{-} + H^{+}$,$i = 1 + \alpha = 1 + 0.06 = 1.06$.
Step $IV$: Calculation of depression in freezing point $(\Delta T_f)$
$\Delta T_f = i \times K_f \times m = 1.06 \times 1.86 \times 0.40 = 0.789 \ K$ (or $^{\circ}C$).

(A) $1$. Calculate the mass of acetic acid $(CH_3COOH)$: $\text{Mass} = \text{Density} \times \text{Volume} = 1.06 \ g \ mL^{-1} \times 0.8 \ mL = 0.848 \ g$.
$2$. Calculate the moles of acetic acid: $\text{Molar mass of } CH_3COOH = 60 \ g \ mol^{-1}$. $\text{Moles} = \frac{0.848 \ g}{60 \ g \ mol^{-1}} \approx 0.01413 \ mol$.
$3$. Calculate the molality $(m)$: Since the solvent is $1 \ kg$ of water,$m = 0.01413 \ mol \ kg^{-1}$.
$4$. Calculate the theoretical depression in freezing point $(\Delta T_f)$: $\Delta T_f = K_f \times m = 1.86 \ K \ kg \ mol^{-1} \times 0.01413 \ mol \ kg^{-1} \approx 0.02628 \ K$.
$5$. Calculate the Van't Hoff factor $(i)$: $i = \frac{\Delta T_f \text{ (observed)}}{\Delta T_f \text{ (theoretical)}} = \frac{0.0325}{0.02628} \approx 1.236 \approx 1.24$.

(C) The dissociation of the complex $A_{x-2}[B(C)_x]_2$ is given by:
$A_{x-2}[B(C)_x]_2 \longrightarrow (x-2) A^{2+} + 2[B(C)_x]^{-(x-2)}$
Total number of ions produced per formula unit,$n = (x-2) + 2 = x$.
The van't Hoff factor $i$ is given by the formula: $i = 1 + (n-1)\alpha$.
Given $i = 4$ and $\alpha = 0.75$:
$4 = 1 + (x-1)(0.75)$
$3 = (x-1)(0.75)$
$x-1 = \frac{3}{0.75} = 4$
$x = 5$.
The coordination number of the central metal ion $B$ is equal to the number of ligands $C$ attached to it,which is $x = 5$.

(B) Statement-$I$: The elevation in boiling point $\Delta T_b$ is given by $\Delta T_b = i \times K_b \times m$.
For $0.1 \ M$ urea (non-electrolyte),the van't Hoff factor $i = 1$.
For $0.1 \ M$ $KCl$ (electrolyte),$KCl$ dissociates as $K^+ + Cl^-$,so $i = 2$.
Since $\Delta T_b \propto i$,the boiling point of $KCl$ solution is higher than that of urea solution. Thus,Statement-$I$ is correct.
Statement-$II$: Elevation of boiling point is a colligative property,which depends on the number of solute particles,not on the molar mass of the solute. Thus,Statement-$II$ is incorrect.

(D) Colligative properties depend on the number of solute particles in the solution.
To determine the highest colligative property,we calculate the van't Hoff factor $(i)$ for each substance:
$1$. $BaCl_2 \rightarrow Ba^{2+} + 2Cl^-$,so $i = 3$.
$2$. $AgNO_3 \rightarrow Ag^+ + NO_3^-$,so $i = 2$.
$3$. $\text{Urea}$ is a non-electrolyte,so $i = 1$.
$4$. $(NH_4)_3PO_4 \rightarrow 3NH_4^+ + PO_4^{3-}$,so $i = 4$.
Since $(NH_4)_3PO_4$ produces the highest number of ions $(i = 4)$,it exhibits the highest colligative properties.

(D) The depression in freezing point is given by the formula $\Delta T_f = i K_f m$.
Since the solutions are equimolal,$K_f$ and $m$ are constant.
Thus,$\Delta T_f \propto i$,where $i$ is the van't Hoff factor.
Freezing point $T_f = T_f^{\circ} - \Delta T_f$.
To have the highest freezing point,the depression in freezing point $\Delta T_f$ must be the minimum,which means $i$ must be the minimum.
For the given substances:
$C_6H_5NH_3Cl \rightarrow C_6H_5NH_3^+ + Cl^-$ $(i = 2)$
$Ba(NO_3)_2 \rightarrow Ba^{2+} + 2NO_3^-$ $(i = 3)$
$LaCl_3 \rightarrow La^{3+} + 3Cl^-$ $(i = 4)$
$C_6H_{12}O_6$ is a non-electrolyte and does not dissociate $(i = 1)$.
Since $C_6H_{12}O_6$ has the lowest $i$ value,it will have the minimum $\Delta T_f$ and consequently the highest freezing point.

(C) The depression in freezing point is given by the formula $\Delta T_f = i \times K_f \times m$.
Since the solutions are equimolal,$m$ is constant.
$K_f$ is also constant for water.
Thus,$\Delta T_f$ depends on the van't Hoff factor $(i)$.
For $BaCl_2$,$i = 3$.
For $Ca(NO_3)_2$,$i = 3$.
For urea,$i = 1$ (as it is a non-electrolyte).
For $Na_2SO_4$,$i = 3$.
Since urea has the minimum value of $i$,it will have the minimum depression in freezing point $(\Delta T_f)$.
Therefore,the freezing point $(T_f = T_f^0 - \Delta T_f)$ will be the highest for urea.

(C) Given,
Mass of sucrose $(w_1) = 17.1 \ g$
Molar mass of sucrose $(M_1) = 342 \ g \ mol^{-1}$
Mass of oxalic acid $(w_2) = x \ g$
Molar mass of oxalic acid $(M_2) = 90 \ g \ mol^{-1}$
Degree of dissociation $(\alpha)$ of oxalic acid $= 0.01$
Step $I$: To find the value of van't Hoff factor $(i)$
$i = (1 - \alpha) + n\alpha$
where,$n = 3$ (number of ions produced by oxalic acid,$H_2C_2O_4 \rightleftharpoons 2H^+ + C_2O_4^{2-}$)
$i = (1 - 0.01) + (3 \times 0.01) = 0.99 + 0.03 = 1.02$
Step $II$: For isotonic solutions,osmotic pressure $(\pi)$ is equal.
$\pi_{\text{sucrose}} = \pi_{\text{oxalic acid}}$
$C_{\text{sucrose}} \times R \times T = i \times C_{\text{oxalic acid}} \times R \times T$
$\frac{w_1}{M_1} = i \times \frac{x}{M_2}$
$x = \frac{w_1 \times M_2}{M_1 \times i} = \frac{17.1 \times 90}{342 \times 1.02} = \frac{1539}{348.84} \approx 4.41$
Thus,$x = 4.41$.

(C) Given:
Mass of solute $(K_3[Fe(CN)_6])$ = $0.1 \ g$
Mass of solvent $(H_2O)$ = $100 \ g$
Molar mass of solute $(M)$ = $329 \ g \ mol^{-1}$
$K_f = 1.86 \ K \ kg \ mol^{-1}$
Dissociation of $K_3[Fe(CN)_6]$:
$K_3[Fe(CN)_6] \rightarrow 3K^{+} + [Fe(CN)_6]^{3-}$
Van't Hoff factor $(i)$ = $3 + 1 = 4$
Molality $(m)$ = $\frac{\text{mass of solute}}{\text{molar mass}} \times \frac{1000}{\text{mass of solvent in } g}$
$m = \frac{0.1}{329} \times \frac{1000}{100} = \frac{1}{329} \ mol \ kg^{-1}$
Depression in freezing point $(\Delta T_f)$:
$\Delta T_f = i \times K_f \times m$
$\Delta T_f = 4 \times 1.86 \times \frac{1}{329} \approx 0.0226 \ K$
Thus,option $C$ is correct.

(B) The formula for elevation in boiling point is $\Delta T_b = i \times K_b \times m$.
Given values are:
$i = 1.98$
$K_b = 0.52 \ K \ kg \ mol^{-1}$
$m = 0.001 \ m$
Substituting these values into the formula:
$\Delta T_b = 1.98 \times 0.52 \times 0.001$
$\Delta T_b = 1.0296 \times 10^{-3} \ K$
Rounding to the appropriate significant figures,we get $\Delta T_b \approx 1.03 \times 10^{-3} \ K$.