Colligative properties of electrolyte Questions in English

(B) The relative lowering of vapour pressure is a colligative property,which depends on the van't Hoff factor $(i)$ and concentration $(C)$: $\Delta P \propto i \times C$.
For $0.1 \ M$ solutions,the concentration $C$ is the same for all.
For $NaCl$,$i = 2$ $(NaCl \rightarrow Na^+ + Cl^-)$.
For $CuSO_4$,$i = 2$ $(CuSO_4 \rightarrow Cu^{2+} + SO_4^{2-})$.
For $K_2SO_4$,$i = 3$ $(K_2SO_4 \rightarrow 2K^+ + SO_4^{2-})$.
Therefore,the ratio is $2 : 2 : 3$,which simplifies to $1 : 1 : 1.5$.

(A) The depression in freezing point is given by $\Delta T_f = i \times K_f \times m$.
For $KCl$,$i = 2$ $(KCl \rightarrow K^+ + Cl^-)$.
For $BaCl_2$,$i = 3$ $(BaCl_2 \rightarrow Ba^{2+} + 2Cl^-)$.
Given $\Delta T_{f(KCl)} = 0 - (-2) = 2^{\circ}C$.
Since $m$ and $K_f$ are the same for both solutions,we have:
$\frac{\Delta T_{f(KCl)}}{i_{KCl}} = \frac{\Delta T_{f(BaCl_2)}}{i_{BaCl_2}}$
$\frac{2}{2} = \frac{\Delta T_{f(BaCl_2)}}{3}$
$\Delta T_{f(BaCl_2)} = 3^{\circ}C$.
Therefore,the freezing point of $BaCl_2$ solution = $0 - 3 = -3^{\circ}C$.

(C) The elevation in boiling point is given by $\Delta T_b = i \times K_b \times m$.
Since the concentration $(m)$ is the same ($1\%$ by mass) for all,$\Delta T_b$ depends on the van't Hoff factor $(i)$.
For glucose,sucrose,and urea,$i = 1$ (non-electrolytes).
For $NaCl$,$i = 2$ (dissociates into $Na^+$ and $Cl^-$).
Since $NaCl$ has the highest $i$ value,it will show the maximum elevation in boiling point,resulting in the highest boiling point.

(C) The dissociation of sodium sulfate is given by: $Na_2SO_4 \rightarrow 2Na^+ + SO_4^{2-}$.
Since it undergoes complete dissociation,the van't Hoff factor $(i)$ is $3$ (two $Na^+$ ions and one $SO_4^{2-}$ ion).
The formula for depression in freezing point is $\Delta T_f = i \times K_f \times m$.
Here,$i = 3$,$K_f = 1.86 \ K \ kg \ mol^{-1}$,and molality $(m)$ = $\frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.01 \ mol}{1 \ kg} = 0.01 \ mol \ kg^{-1}$.
Substituting the values: $\Delta T_f = 3 \times 1.86 \times 0.01 = 0.0558 \ K$.

(B) For $100\%$ ionization,the degree of dissociation $\alpha = 1$.
The complex $K_3[Fe(CN)_6]$ dissociates as:
$K_3[Fe(CN)_6] \rightarrow 3K^+ + [Fe(CN)_6]^{3-}$.
Here,the number of ions produced per formula unit,$n = 3 + 1 = 4$.
The van't Hoff factor $(i)$ is given by the formula:
$i = 1 + (n - 1)\alpha$.
Substituting the values:
$i = 1 + (4 - 1) \times 1 = 4$.

(B) The dissociation of the weak acid is given by: $HX \rightleftharpoons H^+ + X^-$.
Since the degree of dissociation $\alpha = 0.20$,the van't Hoff factor $i = 1 + \alpha(n-1) = 1 + 0.20(2-1) = 1.2$.
The depression in freezing point is calculated as $\Delta T_f = i \times K_f \times m$.
Substituting the values: $\Delta T_f = 1.2 \times 1.86 \times 0.5 = 1.116 \ K \approx 1.12 \ K$.

(C) The freezing point depression is given by $\Delta T_f = i \times K_f \times m$.
For the dissociation of $HX \rightleftharpoons H^+ + X^-$,the van't Hoff factor $i = 1 + \alpha$ (where $n=2$).
Given $\alpha = 0.3$,$i = 1 + 0.3 = 1.3$.
$\Delta T_f = 1.3 \times 1.85 \times 0.2 = 0.481 \ ^oC$.
The freezing point of the solution $T_f = T_f^o - \Delta T_f = 0 - 0.481 \ ^oC = -0.481 \ ^oC$.
Thus,the value is close to $-0.48 \ ^oC$.

(C) The depression in freezing point $(\Delta T_f)$ is a colligative property,which depends on the van't Hoff factor $(i)$.
$\Delta T_f = i \times K_f \times m$.
Since the molality $(m)$ is the same for all solutions,the one with the highest van't Hoff factor $(i)$ will have the greatest depression in freezing point,resulting in the lowest freezing point.
For $NaCl$,$i = 2$ $(Na^+ + Cl^-)$.
For $KCl$,$i = 2$ $(K^+ + Cl^-)$.
For $CaCl_2$,$i = 3$ $(Ca^{2+} + 2Cl^-)$.
For $\text{Urea}$,$i = 1$ (non-electrolyte).
Since $CaCl_2$ has the highest value of $i$ $(i = 3)$,it will show the maximum depression in freezing point,thus having the lowest freezing point.

(A) $1$. Calculate the molar mass of $CaCl_2$: $M = 40 + 2 \times 35.5 = 111 \ g \ mol^{-1}$.
$2$. Calculate the number of moles of $CaCl_2$: $n = \frac{0.0112 \ g}{111 \ g \ mol^{-1}} \approx 1.009 \times 10^{-4} \ mol$.
$3$. Since $CaCl_2$ dissociates as $CaCl_2 \rightarrow Ca^{2+} + 2Cl^-$,the van't Hoff factor $(i)$ is $3$.
$4$. The molality $(m)$ is $\frac{1.009 \times 10^{-4} \ mol}{1 \ kg} = 1.009 \times 10^{-4} \ m$.
$5$. The depression in freezing point is given by $\Delta T_f = i \times K_f \times m$.
$6$. $\Delta T_f = 3 \times 2 \times 1.009 \times 10^{-4} \approx 6.054 \times 10^{-4} \approx 0.0006 \ K$.

(A) The formula for depression in freezing point is $\Delta T_f = i \times K_f \times m$.
First,calculate the molality $(m)$: $m = \frac{\text{mass of solute}}{\text{molar mass}} \times \frac{1000}{\text{mass of solvent in g}} = \frac{0.1}{329} \times \frac{1000}{100} = \frac{1}{329} \approx 0.00304 \ mol \ kg^{-1}$.
$K_3[Fe(CN)_6]$ dissociates as $K_3[Fe(CN)_6] \rightarrow 3K^{+} + [Fe(CN)_6]^{3-}$,so the van't Hoff factor $(i)$ is $4$.
Now,calculate $\Delta T_f$: $\Delta T_f = 4 \times 1.86 \times 0.00304 \approx 0.0226 \ K$.
Since the freezing point of pure water is $0^\circ C$,the freezing point of the solution is $0 - 0.0226 = -0.0226^\circ C$,which is approximately $-2.3 \times 10^{-2} \ ^\circ C$.

(C) For isotonic solutions,the osmotic pressure is equal,so $i_1 C_1 = i_2 C_2$.
For $Na_2SO_4$,the van't Hoff factor $i = 1 + (n - 1)\alpha$,where $n = 3$ $(Na_2SO_4 \rightarrow 2Na^+ + SO_4^{2-})$.
For glucose,$i = 1$ as it is a non-electrolyte.
Substituting the values: $[1 + (3 - 1)\alpha] \times 0.004 = 1 \times 0.01$.
$[1 + 2\alpha] = \frac{0.01}{0.004} = 2.5$.
$2\alpha = 2.5 - 1 = 1.5$.
$\alpha = \frac{1.5}{2} = 0.75$.
Therefore,the degree of dissociation in percentage is $0.75 \times 100 = 75 \%$.

(B) Glucose $(C_6H_{12}O_6)$ is a non-electrolyte solute.
It does not undergo dissociation or association in an aqueous solution.
For non-electrolytes,the van't Hoff factor $(i)$ is equal to $1$.

(A) The osmotic pressure $(\pi)$ is 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 $0.1 \ M$ $KNO_3$ (a strong electrolyte),it dissociates completely into $K^+$ and $NO_3^-$ ions,so $i \approx 2$.
For $0.1 \ M$ $CH_3COOH$ (a weak electrolyte),it dissociates only partially,so $i$ is slightly greater than $1$.
Since $C$,$R$,and $T$ are the same for both solutions,the osmotic pressure depends on the van't Hoff factor $i$.
Because $i_{KNO_3} > i_{CH_3COOH}$,it follows that $P_1 > P_2$.

(B) The elevation in boiling point is a colligative property,which depends on the number of particles in the solution.
$BaCl_2$ dissociates as $BaCl_2 \rightarrow Ba^{2+} + 2Cl^-$,producing $3$ ions per formula unit.
$KCl$ dissociates as $KCl \rightarrow K^+ + Cl^-$,producing $2$ ions per formula unit.
Since the molarity is the same,the van't Hoff factor $(i)$ for $BaCl_2$ $(i \approx 3)$ is greater than that for $KCl$ $(i \approx 2)$.
Therefore,the elevation in boiling point for $BaCl_2$ is greater than that for $KCl$,implying $t_1 > t_2$.

(A) The depression in freezing point $\Delta T_f$ is directly proportional to the van't Hoff factor $i$ for solutions of the same concentration $(C)$.
$\Delta T_f = i \times K_f \times m$.
For $K_2SO_4$,$i = 3$ (as it dissociates into $2K^+ + SO_4^{2-}$).
For $NaCl$,$i = 2$ (as it dissociates into $Na^+ + Cl^-$).
For Urea and Glucose,$i = 1$ (as they are non-electrolytes).
Since $K_2SO_4$ has the highest van't Hoff factor,it will show the maximum depression in freezing point.
Therefore,the $0.1 \, M$ solution of $K_2SO_4$ will have the lowest freezing point.

(C) The depression in freezing point $(\Delta T_f)$ 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,$\Delta T_f$ is directly proportional to the van't Hoff factor $(i)$.
For urea (non-electrolyte),$i = 1$.
For $NaCl$ (dissociates into $Na^+$ and $Cl^-$),$i = 2$.
For $Na_2SO_4$ (dissociates into $2Na^+$ and $SO_4^{2-}$),$i = 3$.
Thus,the ratio of depression in freezing point is $1 : 2 : 3$.

(A) The dissociation of $K_4[Fe(CN)_6]$ is given by: $K_4[Fe(CN)_6] \rightarrow 4K^+ + [Fe(CN)_6]^{4-}$.
Since $5$ ions are produced,the van't Hoff factor $i = 5$.
The elevation in boiling point is calculated as: $\Delta T_b = i \times K_b \times m$.
Substituting the values: $\Delta T_b = 5 \times 0.52 \times 0.01 = 0.026 \ K$.
The boiling point of the solution is $T_b = T_b^\circ + \Delta T_b = 100 + 0.026 = 100.26 \ ^\circ C$.

(B) The dissociation reaction is: $HBr \rightleftharpoons H^{+} + Br^{-}$
Initial moles: $1, 0, 0$
Moles at equilibrium: $(1-\alpha), \alpha, \alpha$
Given degree of ionization $\alpha = 0.9$.
Total moles at equilibrium $= 1 - \alpha + \alpha + \alpha = 1 + \alpha = 1 + 0.9 = 1.9$.
Van't Hoff factor $i = 1.9$.
Molality $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{8.1 / 81}{100 / 1000} = \frac{0.1}{0.1} = 1 \ m$.
Depression in freezing point $\Delta T_f = i \times K_f \times m = 1.9 \times 1.86 \times 1 = 3.534 \ ^\circ C$.
Freezing point of solution $= 0 - \Delta T_f = 0 - 3.534 \approx -3.53 \ ^\circ C$.

(A) The dissociation of $CuCl_2$ is: $CuCl_2 \rightarrow Cu^{2+} + 2Cl^-$.
Since there are $3$ ions produced,the van't Hoff factor $i = 3$ (assuming complete dissociation).
The molality $m$ is calculated as: $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{13.44 \ g / 134.1 \ g \ mol^{-1}}{1 \ kg} \approx 0.1 \ mol \ kg^{-1}$.
The elevation in boiling point is given by $\Delta T_b = i \times K_b \times m$.
$\Delta T_b = 3 \times 0.5 \times 0.1 = 0.15 \ K$.
Rounding to the nearest provided option,the value is $0.16$.

(C) The dissociation of $Ca(NO_3)_2$ in water is represented as:
$Ca(NO_3)_2 \rightarrow Ca^{2+} + 2NO_3^-$
Since one mole of $Ca(NO_3)_2$ produces $1$ mole of $Ca^{2+}$ ions and $2$ moles of $NO_3^-$ ions,the total number of particles produced is $1 + 2 = 3$.
Therefore,the van't Hoff factor $(i)$ is $3$.

(D) The freezing point depression is given by the formula $\Delta T_f = i \times K_f \times m$.
Since the solutions are equimolar ($m$ is constant) and $K_f$ is a constant for the solvent,the freezing point depression depends on the van't Hoff factor $(i)$.
$A$ higher $i$ value leads to a greater depression in the freezing point,which means a lower freezing point.
To have the highest freezing point,the solution must have the lowest $i$ value.
Calculating $i$ for each:
$C_6H_5NH_3Cl \rightarrow C_6H_5NH_3^+ + Cl^-$ $(i = 2)$
$Ca(NO_3)_2 \rightarrow Ca^{2+} + 2NO_3^-$ $(i = 3)$
$La(NO_3)_3 \rightarrow La^{3+} + 3NO_3^-$ $(i = 4)$
$C_6H_{12}O_6$ (Glucose) is a non-electrolyte,so it does not dissociate $(i = 1)$.
Since glucose has the lowest $i$ value $(i = 1)$,it will have the minimum depression in freezing point and thus the highest freezing point.

(B) The lowering of vapor pressure is a colligative property,which depends on the number of particles (van't Hoff factor $i$ $\times$ moles of solute).
For $1\%$ solutions,we calculate the relative number of particles:
$1\%$ Sucrose $(C_{12}H_{22}O_{11})$: $i=1$,moles = $1/342 \approx 0.0029$
$1\% NaCl$: $i=2$,moles = $1/58.5 \approx 0.0342$
$1\% CaCl_2$: $i=3$,moles = $1/111 \approx 0.0090 \times 3 = 0.0270$
$1\%$ Glucose $(C_6H_{12}O_6)$: $i=1$,moles = $1/180 \approx 0.0055$
Since $1\% NaCl$ has the highest number of particles,it will cause the maximum lowering of vapor pressure,resulting in the lowest vapor pressure.

(B) The elevation in boiling point is given by the formula: $\Delta T_b = i \times K_b \times m$.
For $NaCl$,the van't Hoff factor $i = 2$ (since $NaCl$ dissociates into $Na^+$ and $Cl^-$).
Given $K_b = 0.51 \ K \ kg \ mol^{-1}$ and $m = 0.1 \ m$.
$\Delta T_b = 2 \times 0.51 \times 0.1 = 0.102 \ ^oC$.
The boiling point of the solution is $T_b = T_b^0 + \Delta T_b$.
Since the boiling point of pure water $T_b^0 = 100 \ ^oC$,the boiling point of the solution is $100 + 0.102 = 100.102 \ ^oC$,which is approximately $100.1 \ ^oC$.

(D) For the dissociation of weak acid $HX$: $HX \rightleftharpoons H^+_{(aq)} + X^-_{(aq)}$
Initial moles: $1, 0, 0$
Equilibrium moles: $(1 - 0.3), 0.3, 0.3$
Total moles at equilibrium $= 1 - 0.3 + 0.3 + 0.3 = 1.3$
Van't Hoff factor $i = \frac{\text{Total moles at equilibrium}}{\text{Initial moles}} = \frac{1.3}{1} = 1.3$
Depression in freezing point $\Delta T_f = i \times K_f \times m$
$\Delta T_f = 1.3 \times 1.85 \times 0.2 = 0.481 \, ^oC$
Freezing point of solution $= T_f^o - \Delta T_f = 0 \, ^oC - 0.481 \, ^oC = -0.481 \, ^oC \approx -0.48 \, ^oC$.

(C) Colligative properties depend on the number of particles in the solution.
For a sugar solution,which is a non-electrolyte,the van't Hoff factor $(i)$ is $1$.
For a $KCl$ solution,which is a strong electrolyte,it dissociates as $KCl \rightarrow K^+ + Cl^-$,so the van't Hoff factor $(i)$ is $2$.
Since the colligative property is directly proportional to the van't Hoff factor $(i)$,the ratio of the colligative property of $KCl$ to that of sugar is $\frac{i_{KCl}}{i_{sugar}} = \frac{2}{1} = 2$.

(B) For a weak acid $HX$,the dissociation is $HX \rightleftharpoons H^+ + X^-$.
Initial moles: $1, 0, 0$.
Moles at equilibrium: $(1-\alpha), \alpha, \alpha$.
Total moles at equilibrium = $1-\alpha + \alpha + \alpha = 1+\alpha$.
Van't Hoff factor $i = 1+\alpha$.
Given $\alpha = 20\% = 0.2$.
So,$i = 1 + 0.2 = 1.2$.
The depression in freezing point is given by $\Delta T_f = i \times K_f \times m$.
$\Delta T_f = 1.2 \times 1.86 \times 0.2 = 0.4464 \ ^oC$.
Since $\Delta T_f = T_f^0 - T_f$,and $T_f^0 = 0 \ ^oC$ for water,
$0.4464 = 0 - T_f \implies T_f = -0.4464 \ ^oC \approx -0.45 \ ^oC$.

(A) The depression in freezing point is given by $\Delta T_f = i \times K_f \times m$.
Since $K_f$ and $m$ are constant for all solutions,the depression in freezing point depends on the van't Hoff factor $(i)$.
Lower $i$ value leads to a smaller depression in freezing point,which corresponds to a higher freezing point.
$i$ values are:
Urea: $i = 1$
$BaCl_2$: $i = 3$
$KBr$: $i = 2$
$Al_2(SO_4)_3$: $i = 5$
Since urea has the lowest $i$ value,it will have the minimum depression in freezing point and thus the highest freezing point.

(A) For the dissociation of a weak acid: $HX \rightleftharpoons H^+ + X^-$,the number of ions produced $n = 2$.
The degree of dissociation $\alpha$ is given as $20\% = 0.2$.
The van't Hoff factor $i$ is calculated as: $i = 1 + \alpha(n - 1) = 1 + 0.2(2 - 1) = 1.2$.
The depression in freezing point is given by the formula: $\Delta T_f = i \times K_f \times m$.
Substituting the values: $\Delta T_f = 1.2 \times 1.86 \times 0.5$.
$\Delta T_f = 1.116 \approx 1.12 \ K$.

(A) Given: $\Delta T_f = 0.00732 \ ^oC$,$m = 0.0020 \ m$,$K_f = 1.86 \ K \ kg \ mol^{-1}$.
Using the formula for freezing point depression: $\Delta T_f = i \times K_f \times m$.
Substituting the values: $0.00732 = i \times 1.86 \times 0.0020$.
$i = \frac{0.00732}{1.86 \times 0.0020} = \frac{0.00732}{0.00372} = 1.967 \approx 2$.
Therefore,the van't Hoff factor $i$ is $2$.

(B) The van't Hoff factor '$i$' is defined as the ratio of the normal molar mass to the observed molar mass.
For electrolytes,dissociation occurs in the solution,which increases the number of particles.
Since the observed molar mass is inversely proportional to the number of particles,the observed molar mass becomes less than the calculated (normal) molar mass.
Consequently,for dissociation,the value of '$i$' is always greater than $1$.

(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 $Ca(NO_3)_2 \rightarrow Ca^{2+} + 2NO_3^-$,the number of ions $(n') = 3$.
Degree of dissociation $(\alpha) = 0.7$.
Van't Hoff factor $(i) = 1 + (n'-1)\alpha = 1 + (3-1) \times 0.7 = 1 + 1.4 = 2.4$.
Using the formula for relative lowering of vapour pressure: $\frac{P^o - P}{P^o} = i \times X$.
$\frac{760 - P}{760} = 2.4 \times 0.00762 = 0.018288$.
$760 - P = 760 \times 0.018288 \approx 13.9$.
$P = 760 - 13.9 = 746.1 \ mm \ Hg$.

(A) The osmotic pressure $(\pi)$ is a colligative property,which depends on the number of particles in the solution. The formula is $\pi = iCRT$,where $i$ is the van't Hoff factor.
For equimolar solutions,the osmotic pressure is directly proportional to the van't Hoff factor $(i)$.
$1$. $Urea$ is a non-electrolyte,so $i = 1$.
$2$. $BaCl_2$ dissociates as $BaCl_2 \to Ba^{2 } 2Cl^{-}$,so $i = 3$.
$3$. $AlCl_3$ dissociates as $AlCl_3 \to Al^{3 } 3Cl^{-}$,so $i = 4$.
Since the number of particles follows the order $AlCl_3 (4) > BaCl_2 (3) > Urea (1)$,the osmotic pressure follows the same order: $AlCl_3 > BaCl_2 > Urea$.

(D) The depression in freezing point $(\Delta T_f)$ is a colligative property,which is directly proportional to the van't Hoff factor $(i)$,where $\Delta T_f = i \times K_f \times m$.
For equimolal solutions,the freezing point is highest when the depression in freezing point is minimum.
This occurs when the number of particles produced in the solution is minimum.
$C_6H_{12}O_6$ (glucose) is a non-electrolyte,so $i = 1$.
$C_6H_5NH_3^+Cl^-$ dissociates into $2$ ions $(i = 2)$.
$Ca(NO_3)_2$ dissociates into $3$ ions $(i = 3)$.
$La(NO_3)_3$ dissociates into $4$ ions $(i = 4)$.
Since glucose has the lowest $i$ value,it shows the minimum depression in freezing point,resulting in the highest freezing point.

(D) The compound $K_3[Fe(CN)_6]$ dissociates in water as follows:
$K_3[Fe(CN)_6] \to 3K^{+} + [Fe(CN)_6]^{3-}$
Total number of ions produced per formula unit = $3 + 1 = 4$.
Therefore,the Van't Hoff factor $(i)$ is $4$.

(D) Barium hydroxide is a strong electrolyte that dissociates completely in an aqueous solution as follows:
$Ba(OH)_{2} \longrightarrow Ba^{2+} + 2OH^{-}$
Since one mole of $Ba(OH)_{2}$ produces $1$ mole of $Ba^{2+}$ ions and $2$ moles of $OH^{-}$ ions,the total number of particles produced is $1 + 2 = 3$.
The van't Hoff factor $(i)$ is defined as the number of particles produced per formula unit of the electrolyte.
Therefore,$i = 3$.

(B) The van't Hoff factor $(i)$ for a $100 \%$ ionised electrolyte is equal to the total number of ions produced per formula unit.
For $Al_2(SO_4)_3$: $Al_2(SO_4)_3 \rightarrow 2 Al^{3+} + 3 SO_4^{2-}$. Total ions = $2 + 3 = 5$,so $i = 5$.
For $Al(NO_3)_3$: $Al(NO_3)_3 \rightarrow Al^{3+} + 3 NO_3^-$. Total ions = $1 + 3 = 4$,so $i = 4$.
For $K_4[Fe(CN)_6]$: $K_4[Fe(CN)_6] \rightarrow 4 K^{+} + [Fe(CN)_6]^{4-}$. Total ions = $4 + 1 = 5$,so $i = 5$.
For $K_2SO_4$: $K_2SO_4 \rightarrow 2 K^{+} + SO_4^{2-}$. Total ions = $2 + 1 = 3$,so $i = 3$.
For $K_3[Fe(CN)_6]$: $K_3[Fe(CN)_6] \rightarrow 3 K^{+} + [Fe(CN)_6]^{3-}$. Total ions = $3 + 1 = 4$,so $i = 4$.
Thus,$K_4[Fe(CN)_6]$ has the same van't Hoff factor as $Al_2(SO_4)_3$.

(C) The depression in freezing point $(\Delta T_f)$ is given by the formula $\Delta T_f = i \times K_f \times m$,where $i$ is the van't Hoff factor,$K_f$ is the cryoscopic constant,and $m$ is the molality.
Since $K_f$ and $m$ are constant for all solutions,$\Delta T_f \propto i$.
We calculate the van't Hoff factor $(i)$ for each solute:
$1. KCl \rightarrow K^+ + Cl^-$,so $i = 2$.
$2. C_6H_{12}O_6$ (glucose) is a non-electrolyte,so $i = 1$.
$3. Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-}$,so $i = 5$.
$4. K_2SO_4 \rightarrow 2K^+ + SO_4^{2-}$,so $i = 3$.
Since $Al_2(SO_4)_3$ has the largest van't Hoff factor $(i = 5)$,it will exhibit the largest freezing point depression.

(B) The formula for freezing point depression is $\Delta T_f = i \times K_f \times m.$
Here,$\Delta T_f = 3.82 \, ^oC,$ $K_f = 1.86 \, ^oC \, kg \, mol^{-1},$ $W_{solute} = 5.00 \, g,$ $W_{solvent} = 45.0 \, g,$ and the molar mass of $Na_2SO_4$ is $M = 142 \, g \, mol^{-1}.$
The molality $m$ is given by $\frac{W_{solute} \times 1000}{M \times W_{solvent}} = \frac{5.00 \times 1000}{142 \times 45.0} \approx 0.782 \, m.$
Substituting these values into the equation: $3.82 = i \times 1.86 \times 0.782.$
Solving for $i$: $i = \frac{3.82}{1.86 \times 0.782} \approx 2.63.$

(D) For the dissociation of weak acid $HX$: $HX \rightleftharpoons H^+ + X^-$
Initial moles: $1 \quad 0 \quad 0$
At equilibrium: $(1-\alpha) \quad \alpha \quad \alpha$
Total moles at equilibrium $= 1 - \alpha + \alpha + \alpha = 1 + \alpha$
Van't Hoff factor $i = 1 + \alpha = 1 + 0.3 = 1.3$
Depression in freezing point $\Delta T_f = i \times K_f \times m$
$\Delta T_f = 1.3 \times 1.86 \times 0.1 = 0.2418 \ ^{\circ}C$
Freezing point of solution $T_f = T_f^{\circ} - \Delta T_f = 0 - 0.2418 \ ^{\circ}C = - 0.24 \ ^{\circ}C$

(D) The depression in freezing point is given by the formula: $\Delta T_f = i \cdot K_f \cdot m$
Here,$\Delta T_f = 0 - (-0.00732) = 0.00732 \ ^oC$,$K_f = 1.86 \ ^oC/m$,and $m = 0.0020 \ m$.
Calculating the van't Hoff factor $(i)$:
$i = \frac{\Delta T_f}{K_f \cdot m} = \frac{0.00732}{1.86 \times 0.0020} = \frac{0.00732}{0.00372} \approx 1.968 \approx 2$.
The ionic compound $[Co(NH_3)_5(NO_2)]Cl$ dissociates in water as: $[Co(NH_3)_5(NO_2)]Cl \rightarrow [Co(NH_3)_5(NO_2)]^+ + Cl^-$.
Thus,$1 \ mol$ of the compound produces $2 \ mol$ of ions.

(B) The depression in freezing point is given by the formula: $\Delta T_{f} = i \times K_{f} \times m$.
For the dissociation of weak acid $HX \rightleftharpoons H^{+} + X^{-}$,the van't Hoff factor $i$ is calculated as $i = 1 + \alpha$,where $\alpha$ is the degree of dissociation.
Given $\alpha = 20\% = 0.2$,so $i = 1 + 0.2 = 1.2$.
Given molality $m = 0.5\, mol\, kg^{-1}$ and $K_{f} = 1.86\, K\, kg\, mol^{-1}$.
Substituting these values: $\Delta T_{f} = 1.2 \times 1.86 \times 0.5 = 1.116\, K \approx 1.12\, K$.

(B) Sodium sulphate $(Na_2SO_4)$ dissociates in aqueous solution as follows:
$Na_2SO_4(aq) \rightarrow 2Na^+(aq) + SO_4^{2-}(aq)$
Since it dissociates into $3$ ions,the van't Hoff factor $(i)$ is $3$.
Given:
Molality $(m) = \frac{0.01 \ mol}{1 \ kg} = 0.01 \ mol \ kg^{-1}$
$K_f = 1.86 \ K \ kg \ mol^{-1}$
The depression in freezing point is given by the formula:
$\Delta T_f = i \times K_f \times m$
$\Delta T_f = 3 \times 1.86 \times 0.01 = 0.0558 \ K$

(A) For the weak acid $HX$,the dissociation is $HX \rightleftharpoons H^+ + X^-$. Here,$n = 2$.
The degree of dissociation $\alpha = 0.20$.
The van't Hoff factor $i = 1 + \alpha(n - 1) = 1 + 0.20(2 - 1) = 1.2$.
The depression in freezing point is given by $\Delta T_f = i \times K_f \times m$.
$\Delta T_f = 1.2 \times 1.86 \, K \cdot kg \cdot mol^{-1} \times 0.2 \, mol \cdot kg^{-1} = 0.4464 \, K$.
Since the freezing point of pure water is $0 \, ^oC$,the freezing point of the solution is $0 - 0.4464 = -0.4464 \, ^oC \approx -0.45 \, ^oC$.

(B) Osmotic pressure is given by the formula $\Pi = iCRT$,where $i$ is the van't Hoff factor,$C$ is molarity,$R$ is the gas constant,and $T$ is temperature.
Since both solutions have the same molarity $(0.1 \ M)$ and are at the same temperature,the osmotic pressure depends on the van't Hoff factor $(i)$.
$KNO_3$ is a strong electrolyte and dissociates completely into two ions ($K^+$ and $NO_3^-$),so its $i \approx 2$.
$CH_3COOH$ is a weak electrolyte and dissociates only to a small extent,so its $i$ is slightly greater than $1$.
Since $i_{KNO_3} > i_{CH_3COOH}$,it follows that $P_1 > P_2$.

(B) Given: $\pi = 16.42 \ atm$,$T = 300 \ K$,$R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$.
Concentration $C = \frac{4.44 \ g}{111 \ g/mol} \times \frac{1000 \ mL}{100 \ mL} = 0.4 \ M$.
Using the formula $\pi = iCRT$,we find the van't Hoff factor $i = \frac{\pi}{CRT} = \frac{16.42}{0.4 \times 0.082 \times 300} = \frac{16.42}{9.84} \approx 1.668 \approx 1.67$.
For $CaCl_2 \longrightarrow Ca^{2+} + 2Cl^-$,the number of ions $n = 3$.
The degree of dissociation $\alpha$ is given by $i = 1 + (n-1)\alpha$.
$1.67 = 1 + (3-1)\alpha \implies 0.67 = 2\alpha \implies \alpha = 0.335$.
Thus,the degree of dissociation is approximately $33.5 \%$,which corresponds to $33 \%$.

(C) $1$. Determine the empirical formula: The molar ratios are $Na : Cu : F = 1.08 : 0.539 : 2.16$. Dividing by $0.539$,we get $Na : Cu : F \approx 2 : 1 : 4$. Thus,the empirical formula is $Na_2CuF_4$.
$2$. Determine the van't Hoff factor $(i)$: The osmotic pressure is three times that of urea (a non-electrolyte,$i=1$),so $i = 3$.
$3$. Analyze the dissociation: $A$ compound with $i=3$ dissociates into $3$ ions. $Na_2[CuF_4]$ dissociates as $Na_2[CuF_4] \rightarrow 2Na^+ + [CuF_4]^{2-}$. This gives $2 + 1 = 3$ ions.
$4$. Conclusion: The formula is $Na_2[CuF_4]$.

(B) Given: $P^0 = 76 \ mm \ Hg$,$P_s = 38 \ mm \ Hg$.
Moles of water $(n_A)$ = $\frac{36 \ g}{18 \ g/mol} = 2 \ mol$.
Moles of solute $(n_B)$ = $1 \ mol$.
Using Raoult's law for non-volatile solute: $\frac{P^0 - P_s}{P_s} = i \times \frac{n_B}{n_A}$.
$\frac{76 - 38}{38} = i \times \frac{1}{2} \implies 1 = i \times \frac{1}{2} \implies i = 2$.
For $Na_2SO_4 \rightarrow 2Na^+ + SO_4^{2-}$,the number of ions produced per formula unit is $n = 3$.
$i = 1 + (n - 1)\alpha \implies 2 = 1 + (3 - 1)\alpha$.
$2 = 1 + 2\alpha \implies 2\alpha = 1 \implies \alpha = 0.5$.
Percentage dissociation = $0.5 \times 100 = 50\%$.

(A) For $BaCl_2$,the dissociation reaction is: $BaCl_2 \rightarrow Ba^{2+} + 2Cl^-$.
Since the salt undergoes $100 \%$ ionisation,the van't Hoff factor $(i)$ is $1 + 2 = 3$.
The depression in freezing point is calculated using the formula: $\Delta T_f = i \times K_f \times m$.
Substituting the values: $\Delta T_f = 3 \times 1.86 \ K \ kg \ mol^{-1} \times 0.05 \ mol \ kg^{-1} = 0.279 \ K$.
The freezing point of the solution $(T_f)$ is given by: $T_f = T_f^0 - \Delta T_f = 0 \ ^oC - 0.279 \ ^oC = -0.279 \ ^oC$.

(B) The dissociation reaction for $Hg_{2}Cl_{2}$ is: $Hg_{2}Cl_{2} \rightleftharpoons Hg_{2}^{2+} + 2Cl^{-}$
The number of ions produced per formula unit is $n = 3$.
The degree of ionization is $\alpha = 80\% = 0.8$.
The formula for the Van't Hoff factor $(i)$ is: $i = 1 + (n - 1)\alpha$.
Substituting the values: $i = 1 + (3 - 1) \times 0.8$.
$i = 1 + 2 \times 0.8 = 1 + 1.6 = 2.6$.