Colligative properties of electrolyte Questions in English

(C) Interionic attractions occur when an ion is surrounded by an ionic atmosphere that has a net charge opposite to its own.
For example,an anion is surrounded by cations,and a cation is surrounded by anions.
Therefore,the strength of the interionic attraction depends directly on the interaction of the charges present on the ions.

(C) $C_{12}H_{22}O_{11}$ (sucrose) is a covalent compound that does not dissociate into ions in an aqueous solution.
Therefore,it is a non-electrolyte.

(C) . Calcium nitrate $(Ca(NO_3)_2)$ is an ionic salt that dissociates completely into $Ca^{2+}$ and $NO_3^-$ ions when dissolved in water.
Sucrose,sulphur in $CS_2$,and ethanol are non-electrolytes and do not produce ions in their respective solutions.

(C) The addition of a polar solvent to a solid electrolyte results in $Ionization$.
Polar solvents have a high dielectric constant,which reduces the electrostatic force of attraction between the ions of the solid electrolyte,thereby facilitating their dissociation into free ions.

(B) When $HgI_2$ is added to an aqueous solution of $KI$,it reacts to form a soluble complex: $HgI_2 + 2KI \rightarrow K_2[HgI_4]$.
This reaction increases the total number of solute particles in the solution.
According to the colligative property,the depression of freezing point $(\Delta T_f)$ is directly proportional to the van't Hoff factor $(i)$ and the molality of the solution.
Since the number of particles increases,the freezing point of the solution decreases.

$(A)$ Raoult's law is applicable to ideal solutions containing non-volatile, non-electrolyte solutes.
It is not applicable if the solute undergoes association or dissociation in the solution, as this changes the total number of particles.
Urea, glucose, and sucrose are non-electrolytes and do not dissociate in water.
$NaCl$ is an electrolyte that dissociates into $Na^{+}$ and $Cl^{-}$ ions in solution.
Therefore, Raoult's law is not applicable to $1 \ M \ NaCl$.

(B) For $KCl$,the van't Hoff factor $i = 2$ because it dissociates as $KCl \rightarrow K^+ + Cl^-$.
The formula for elevation in boiling point is $\Delta T_b = i \times K_b \times m$.
Given $i = 2$,$K_b = 0.52 \, ^{\circ}C \, kg \, mol^{-1}$,and $m = 1.0 \, m$.
$\Delta T_b = 2 \times 0.52 \times 1.0 = 1.04 \, ^{\circ}C$.
The boiling point of the solution $T_b = T_b^{\circ} + \Delta T_b$.
Since the boiling point of pure water $T_b^{\circ} = 100 \, ^{\circ}C$,we have $T_b = 100 + 1.04 = 101.04 \, ^{\circ}C$.

(B) Elevation in boiling point $(\Delta T_b)$ is a colligative property,which depends on the number of particles in the solution.
The formula is $\Delta T_b = i \times K_b \times m$,where $i$ is the van't Hoff factor.
For sucrose (a non-electrolyte),$i = 1$. Thus,$\Delta T_b(\text{sucrose}) = 1 \times K_b \times m = 0.1\,^{\circ}C$.
For $NaCl$ (a strong electrolyte),it dissociates as $NaCl \rightarrow Na^+ + Cl^-$,so $i = 2$.
Thus,$\Delta T_b(NaCl) = 2 \times K_b \times m = 2 \times (K_b \times m) = 2 \times 0.1\,^{\circ}C = 0.2\,^{\circ}C$.

(A) Step $1$: Calculate the number of moles of $CuCl_2$. $\text{Moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{13.44 \ g}{134.4 \ g \ mol^{-1}} = 0.1 \ mol$.
Step $2$: Calculate molality $(m)$. $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.1 \ mol}{1 \ kg} = 0.1 \ m$.
Step $3$: Determine the van't Hoff factor $(i)$. For $CuCl_2$,the dissociation is $CuCl_2 \to Cu^{2+} + 2Cl^-$. Assuming $100 \%$ ionization,$i = 3$.
Step $4$: Calculate elevation in boiling point $(\Delta T_b)$. $\Delta T_b = i \times K_b \times m = 3 \times 0.52 \times 0.1 = 0.156 \ K \approx 0.16 \ K$.

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

(B) The molar mass of $HBr = 1 + 80 = 81 \ g \ mol^{-1}$.
Number of moles of $HBr = \frac{8.1 \ g}{81 \ g \ mol^{-1}} = 0.1 \ mol$.
Molality $(m) = \frac{0.1 \ mol}{0.1 \ kg} = 1 \ m$.
For $HBr \rightleftharpoons H^{+} + Br^{-}$,the degree of dissociation $\alpha = 0.9$.
Van't Hoff factor $(i) = 1 + \alpha = 1 + 0.9 = 1.9$.
Depression in freezing point $\Delta T_f = i \times K_f \times m = 1.9 \times 1.86 \times 1 = 3.534 \ K$.
Freezing point of solution $T_f = T_f^{\circ} - \Delta T_f = 0 - 3.534 = - 3.534 \ ^oC$.

(B) For $NaCl$,the van't Hoff factor $i = 2$ because it dissociates as $NaCl \rightarrow Na^+ + Cl^-$.
The formula for depression in freezing point is $\Delta T_f = i \times K_f \times m$.
Given $i = 2$,$K_f = 1.86$,and $m = 1$,we have $\Delta T_f = 2 \times 1.86 \times 1 = 3.72$.
The freezing point of the solution is $T_f = T_f^{\circ} - \Delta T_f = 0 - 3.72 = - 3.72 \, ^oC$.

(C) The osmotic pressure $(O.P.)$ is a colligative property,which depends on the van't Hoff factor $(i)$.
$O.P. = i \times C \times R \times T$.
For a given concentration $(1 \ M)$,$O.P.$ is directly proportional to the number of ions produced in the solution $(i)$.
$AgNO_3 \rightarrow Ag^+ + NO_3^-$ $(i = 2)$
$MgCl_2 \rightarrow Mg^{2+} + 2Cl^-$ $(i = 3)$
$(NH_4)_3PO_4 \rightarrow 3NH_4^+ + PO_4^{3-}$ $(i = 4)$
$Na_2SO_4 \rightarrow 2Na^+ + SO_4^{2-}$ $(i = 3)$
Since $(NH_4)_3PO_4$ produces the maximum number of ions $(i = 4)$,it will show the maximum osmotic pressure.

(B) The lowering of vapour pressure is a colligative property,which depends on the number of particles (van't Hoff factor,$i$) in the solution.
$BaCl_2$ dissociates as $BaCl_2 \rightarrow Ba^{2+} + 2Cl^-$,giving $3$ ions per formula unit $(i = 3)$.
$NaCl$ dissociates as $NaCl \rightarrow Na^+ + Cl^-$,giving $2$ ions per formula unit $(i = 2)$.
$KCl$ dissociates as $KCl \rightarrow K^+ + Cl^-$,giving $2$ ions per formula unit $(i = 2)$.
Since $BaCl_2$ produces the highest number of particles,it causes the greatest lowering of vapour pressure,resulting in the lowest vapour pressure among the given options.

(C) 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 $NaCl$,it dissociates as $NaCl \rightarrow Na^+ + Cl^-$,so the van't Hoff factor $i = 2$.
For $Na_2SO_4$,it dissociates as $Na_2SO_4 \rightarrow 2Na^+ + SO_4^{2-}$,so the van't Hoff factor $i = 3$.
Since both solutions have the same molar concentration $(C = 0.1 \ M)$ and are at the same temperature,the osmotic pressure depends directly on the van't Hoff factor $i$.
Because $i$ for $Na_2SO_4$ $(i=3)$ is greater than $i$ for $NaCl$ $(i=2)$,the osmotic pressure of the $Na_2SO_4$ solution will be higher.

(C) 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 glucose,it is a non-electrolyte,so $i = 1$.
For $NaCl$,it dissociates as $NaCl \rightarrow Na^+ + Cl^-$,so $i = 2$.
For $BaCl_2$,it dissociates as $BaCl_2 \rightarrow Ba^{2+} + 2Cl^-$,so $i = 3$.
Since the solutions are equimolar ($C$ is constant),the osmotic pressure depends directly on $i$. Therefore,the order is $BaCl_2 (i=3) > NaCl (i=2) >$ Glucose $(i=1)$.

(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.
Since $C$,$R$,and $T$ are constant for all given solutions,the osmotic pressure depends directly on the van't Hoff factor $(i)$.
For $CaCl_2$,$i = 3$ $(Ca^{2+} + 2Cl^-)$.
For $KCl$,$i = 2$ $(K^+ + Cl^-)$.
For Glucose and Urea,$i = 1$ because they are non-electrolytes.
Since $CaCl_2$ has the highest value of $i$,it exhibits the highest osmotic pressure.

(C) The elevation in boiling point is a colligative property,which depends on the van't Hoff factor $(i)$.
$i$ for $NaCl$ is $2$ $(Na^+ + Cl^-)$.
$i$ for sucrose is $1$ (non-electrolyte).
$i$ for $BaCl_2$ is $3$ $(Ba^{2+} + 2Cl^-)$.
$i$ for glucose is $1$ (non-electrolyte).
Since $BaCl_2$ has the highest van't Hoff factor,it will produce the maximum number of ions in the solution,leading to the highest elevation in boiling point.

(D) The elevation in boiling point is a colligative property,which depends on the number of particles in the solution.
$1\%$ solutions imply equal mass concentrations.
Glucose and sucrose are non-electrolytes $(i = 1)$.
$NaCl$ dissociates into $2$ ions $(Na^+ + Cl^-)$,so $i = 2$.
$CaCl_2$ dissociates into $3$ ions $(Ca^{2+} + 2Cl^-)$,so $i = 3$.
Since $CaCl_2$ produces the maximum number of particles per unit mass,it exhibits the highest boiling point elevation.
Therefore,the correct option is $(D)$.

(D) The boiling point elevation is a colligative property,which depends on the van't Hoff factor $(i)$.
$KCl$ is a strong electrolyte that dissociates completely into $K^+$ and $Cl^-$ ions $(i \approx 2)$.
Phenol $(C_6H_5OH)$ is a weak acid that undergoes partial dissociation in water,providing more particles than a non-electrolyte $(i > 1)$.
Glucose $(C_6H_{12}O_6)$ is a non-electrolyte and does not dissociate $(i = 1)$.
Since the boiling point elevation $\Delta T_b = i \times K_b \times m$,the order of boiling points is $KCl > C_6H_5OH > C_6H_{12}O_6$.

(C) The boiling point elevation is a colligative property,which depends on the van't Hoff factor $(i)$.
For $0.1 \ N \ Na_2SO_4$,$i = 3$ and molarity $= 0.05 \ M$,so $i \times M = 0.15$.
For $0.1 \ N \ MgSO_4$,$i = 2$ and molarity $= 0.05 \ M$,so $i \times M = 0.10$.
For $0.1 \ M \ Al_2(SO_4)_3$,$i = 5$ and molarity $= 0.1 \ M$,so $i \times M = 0.50$.
For $0.1 \ M \ BaSO_4$,it is sparingly soluble,so $i \times M$ is very low.
Since $Al_2(SO_4)_3$ has the highest value of $i \times M$,it will show the highest boiling point.

(C) The depression of 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.
Since the molality $(m)$ and the freezing point depression constant $(K_f)$ are the same for all solutions, the ratio of $\Delta T_f$ is equal to the ratio of their van't Hoff factors $(i)$.
For urea (a non-electrolyte), $i = 1$.
For common salt $(NaCl)$, it dissociates as $NaCl \rightarrow Na^+ + Cl^-$, so $i = 2$.
For $Na_2SO_4$, it dissociates as $Na_2SO_4 \rightarrow 2Na^+ + SO_4^{2-}$, so $i = 3$.
Therefore, the ratio of depression of freezing point is $1 : 2 : 3$.

(C) The depression in freezing point is a colligative property,which depends on the van't Hoff factor $(i)$.
$i$ for $NaCl$ is $2$ $(Na^+ + Cl^-)$.
$i$ for $KCl$ is $2$ $(K^+ + Cl^-)$.
$i$ for $CaCl_2$ is $3$ $(Ca^{2+} + 2Cl^-)$.
$i$ for urea is $1$ (non-electrolyte).
Since the depression in freezing point $\Delta T_f = i \times K_f \times m$,the solution with the highest $i$ value will have the maximum depression in freezing point,resulting in the minimum freezing point.
Therefore,$CaCl_2$ has the minimum freezing point.

(B) The freezing point depression 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 freezing point depends on the van't Hoff factor $(i)$.
Lower freezing point corresponds to a higher value of $i$.
For glucose and urea,$i = 1$.
For $NaCl$,$i = 2$ $(NaCl \rightarrow Na^+ + Cl^-)$.
For $ZnSO_4$,$i = 2$ $(ZnSO_4 \rightarrow Zn^{2+} + SO_4^{2-})$.
However,$NaCl$ is a strong electrolyte with a higher degree of dissociation compared to $ZnSO_4$ in this concentration range,leading to a greater depression in freezing point.
Thus,the $0.1 \ M$ aqueous solution of $NaCl$ has the lowest freezing point.

(A) Depression in freezing point $(\Delta T_f)$ is a colligative property,which is directly proportional to the van't Hoff factor $(i)$ for equimolar solutions: $\Delta T_f = i \times K_f \times m$.
For the given solutes:
$i (\text{Glucose}) = 1$ (non-electrolyte)
$i (KNO_3) = 2$ $(K^{+} + NO_3^{-})$
$i (AlCl_3) = 4$ $(Al^{3+} + 3Cl^{-})$
Since $\Delta T_f \propto i$,the order of depression in freezing point is: $\text{Glucose} < KNO_3 < AlCl_3$.
Freezing point $(T_f)$ is related to depression as $T_f = T_f^{\circ} - \Delta T_f$.
Therefore,a higher depression leads to a lower freezing point.
The correct order of freezing points is: $AlCl_3 < KNO_3 < \text{Glucose}$.

(A) The freezing point depression is given by the formula $\Delta T_f = i \times K_f \times m$.
Since the molality $(m)$ and the solvent $(K_f)$ are the same for all solutions,the freezing point depends on the van't Hoff factor $(i)$.
$Al_2(SO_4)_3$ dissociates as $2Al^{3+} + 3SO_4^{2-}$,giving $i = 5$.
$C_5H_{10}O_5$ is a non-electrolyte,$i = 1$.
$KI$ dissociates as $K^+ + I^-$,giving $i = 2$.
$C_{12}H_{22}O_{11}$ is a non-electrolyte,$i = 1$.
Since $Al_2(SO_4)_3$ has the highest van't Hoff factor $(i=5)$,it will show the maximum depression in freezing point,resulting in the lowest freezing point.

(B) Colligative properties are directly proportional to the van't Hoff factor $(i)$ for solutions of the same concentration.
$1.$ For $NaCl$: $NaCl \to Na^{+} + Cl^{-}$,so $i = 2$.
$2.$ For $Na_2SO_4$: $Na_2SO_4 \to 2Na^{+} + SO_4^{2-}$,so $i = 3$.
$3.$ For $Na_3PO_4$: $Na_3PO_4 \to 3Na^{+} + PO_4^{3-}$,so $i = 4$.
Since the colligative property depends on the number of particles,the order is $NaCl < Na_2SO_4 < Na_3PO_4$ $(2 < 3 < 4)$.

(C) The Van't Hoff factor $(i)$ is equal to the number of ions produced upon dissociation of the solute in the solution.
$A$. $NaCl \rightarrow Na^+ + Cl^-$ $(i = 2)$
$B$. $MgCl_2 \rightarrow Mg^{2+} + 2Cl^-$ $(i = 3)$
$C$. $Na_3PO_4 \rightarrow 3Na^+ + PO_4^{3-}$ $(i = 4)$
$D$. $Urea$ $(NH_2CONH_2)$ is a non-electrolyte,so $i = 1$.
Since $Na_3PO_4$ produces the maximum number of ions $(4)$,it has the highest Van't Hoff factor.

(A) The Van't Hoff factor $i$ represents the number of particles a solute dissociates into in a solution.
For $K_4[Fe(CN)_6]$,the dissociation is: $K_4[Fe(CN)_6] \rightarrow 4K^{+} + [Fe(CN)_6]^{4-}$.
Total number of particles $i = 4 + 1 = 5$.
Now,checking the options:
$A$. $Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-}$,so $i = 2 + 3 = 5$.
$B$. $NaCl \rightarrow Na^{+} + Cl^{-}$,so $i = 1 + 1 = 2$.
$C$. $Na_2SO_4 \rightarrow 2Na^{+} + SO_4^{2-}$,so $i = 2 + 1 = 3$.
$D$. $Al(NO_3)_3 \rightarrow Al^{3+} + 3NO_3^{-}$,so $i = 1 + 3 = 4$.
Therefore,$Al_2(SO_4)_3$ has the same Van't Hoff factor as $K_4[Fe(CN)_6]$.

(C) Colligative properties depend on the number of particles in the solution.
For $KCl$ (a strong electrolyte),it dissociates as $KCl \rightarrow K^+ + Cl^-$,so the Van't Hoff factor $(i)$ is $2$.
For sugar (a non-electrolyte),it does not dissociate,so the Van't Hoff factor $(i)$ is $1$.
The ratio of the colligative property of $KCl$ solution to that of sugar solution is equal to the ratio of their Van't Hoff factors,which is $\frac{2}{1} = 2.0$.

(C) $Ca(NO_3)_2$ dissociates in aqueous solution as follows:
$Ca(NO_3)_2 \to Ca^{2+} + 2NO_3^-$
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 ions produced is $1 + 2 = 3$.
Therefore,the Van't Hoff factor $(i)$ is $3$.

(D) The Van't Hoff factor $i$ represents the number of particles a solute dissociates into in a solution.
For $K_2SO_4$,$i = 3$ $(2K^+ + SO_4^{2-})$.
For $NaHSO_4$,$i = 2$ ($Na^+ + H^+ + SO_4^{2-}$ is incorrect,it dissociates as $Na^+ + HSO_4^-$ in dilute solution,so $i = 2$).
For Sugar,$i = 1$ (non-electrolyte).
For $MgSO_4$,$i = 2$ $(Mg^{2+} + SO_4^{2-})$.
Both $NaHSO_4$ and $MgSO_4$ have $i = 2$. However,$MgSO_4$ is the standard example for a $1:1$ electrolyte dissociating into $2$ ions.

(B) The Van't Hoff factor $i$ is defined as the ratio of the calculated molecular weight to the experimental molecular weight.
For electrolytes,dissociation occurs in solution,which increases the number of particles.
Since $i = \frac{\text{Calculated molecular weight}}{\text{Experimental molecular weight}}$,and for dissociation $i > 1$,the experimental molecular weight is always less than the calculated molecular weight.

(B) The boiling point elevation is a colligative property,which depends on the van't Hoff factor $(i)$.
For $AlCl_3$,the dissociation is $AlCl_3 \rightarrow Al^{3+} + 3Cl^-$,so $i = 4$.
For $CaCl_2$,the dissociation is $CaCl_2 \rightarrow Ca^{2+} + 2Cl^-$,so $i = 3$.
Since the molar concentration is the same,the boiling point elevation $\Delta T_b = i \times K_b \times m$ is directly proportional to $i$.
Because $i$ for $AlCl_3$ $(4)$ is greater than $i$ for $CaCl_2$ $(3)$,the boiling point $t_1$ of $AlCl_3$ will be greater than $t_2$ of $CaCl_2$,i.e.,$t_1 > t_2$.

(D) Sodium phosphate $(Na_3PO_4)$ is a strong electrolyte that dissociates completely in water as follows:
$Na_3PO_4 \rightarrow 3Na^{+} + PO_4^{3-}$
Since $1$ mole of $Na_3PO_4$ produces $3$ moles of $Na^{+}$ ions and $1$ mole of $PO_4^{3-}$ ion,the total number of particles produced is $3 + 1 = 4$.
Therefore,the Van't Hoff factor $(i)$ is $4$.

(B) For a weak acid $HX$ undergoing dissociation: $HX \rightleftharpoons H^+ + X^-$.
The van't Hoff factor $i = 1 + \alpha$,where $\alpha$ is the degree of dissociation.
Given $\alpha = 20 \% = 0.2$,so $i = 1 + 0.2 = 1.2$.
The depression in freezing point is given by $\Delta T_f = i \times K_f \times m$.
Substituting the values: $\Delta T_f = 1.2 \times 1.86 \, ^o C/m \times 0.2 \, m = 0.4464 \, ^o C$.
The freezing point of the solution is $T_f = T_f^o - \Delta T_f = 0 \, ^o C - 0.4464 \, ^o C = - 0.4464 \, ^o C \approx - 0.45 \, ^o C$.

(B) The formula for freezing point depression is $\Delta T_f = i \times K_f \times m$.
Given $\Delta T_f = 0.0054 \ ^oC$,$K_f = 1.80 \ K \ kg \ mol^{-1}$,and $m = 0.001 \ molal$.
Substituting the values: $0.0054 = i \times 1.80 \times 0.001$.
$i = \frac{0.0054}{0.0018} = 3$.
Since the van't Hoff factor $i = 3$,the complex must dissociate into $3$ ions in solution.
The complex $[Pt(NH_3)_4Cl_2]Cl_2$ dissociates as $[Pt(NH_3)_4Cl_2]^{2+} + 2Cl^-$,giving a total of $3$ ions.
Therefore,the correct formulation is $[Pt(NH_3)_4Cl_2]Cl_2$.

(C) When a polar solvent is added to a solid electrolyte,the electrostatic forces of attraction between the ions are weakened due to the high dielectric constant of the solvent,resulting in the dissociation of the electrolyte into its constituent ions. This process is known as $Ionization$.

(C) non-electrolyte is a substance that does not dissociate into ions in an aqueous solution and therefore does not conduct electricity.
$NaCl$,$CaCl_2$,and $CH_3COOH$ are electrolytes because they dissociate into ions in water.
$C_{12}H_{22}O_{11}$ (sucrose) is a covalent compound that dissolves in water as molecules without dissociating into ions,making it a non-electrolyte.
Therefore,the correct option is $C$.

(A) The substances whose aqueous solutions allow the passage of electric current and are chemically decomposed are termed electrolytes.
Electrolytic substances are classified as strong or weak according to how readily they dissociate into conducting ions.
Acetic acid $(CH_3COOH)$ is a weak electrolyte because it partially dissociates into ions in an aqueous solution.
Glucose $(C_6H_{12}O_6)$,ethanol $(C_2H_5OH)$,and urea $(NH_2CONH_2)$ are non-electrolytes as they do not dissociate into ions in an aqueous solution.

(A) Carnallite is a double salt with the molecular formula $KCl \cdot MgCl_2 \cdot 6H_2O$.
When dissolved in $H_2O$,it completely dissociates into its constituent ions.
Therefore,the solution contains $K^{+}$,$Mg^{2+}$,and $Cl^{-}$ ions.

(B) The formula for freezing point depression is $\Delta T_f = i \times K_f \times m$.
Given $\Delta T_f = 0.0054$,$K_f = 1.80$,and $m = 0.01$.
Substituting the values: $0.0054 = i \times 1.80 \times 0.01$.
$i = \frac{0.0054}{0.018} = 3$.
Since the van't Hoff factor $i = 3$,the complex dissociates into $3$ ions in the solution.
Among the options,$[Pt(NH_3)_4Cl_2]Cl_2$ dissociates as $[Pt(NH_3)_4Cl_2]^{2+} + 2Cl^-$,which gives $1 + 2 = 3$ ions.
Therefore,the correct formula is $[Pt(NH_3)_4Cl_2]Cl_2$.

(D) The depression in freezing point is a colligative property,which depends on the van't Hoff factor $(i)$.
For non-electrolytes like Urea and Glucose,$i = 1$.
For $NaCl$,$i = 2$ $(NaCl \rightarrow Na^+ + Cl^-)$.
For $K_2SO_4$,$i = 3$ $(K_2SO_4 \rightarrow 2K^+ + SO_4^{2-})$.
Since $K_2SO_4$ produces the highest number of ions $(i = 3)$,it will show the highest depression in freezing point.

(B) The freezing point depression is given by the formula $\Delta T_f = i \times K_f \times m$.
Since $KCl$ dissociates completely as $KCl \rightarrow K^+ + Cl^-$,the van't Hoff factor $i = 2$.
Given molality $m = 1 \ m$ and $K_f = 1.86 \ K \ kg \ mol^{-1}$.
$\Delta T_f = 2 \times 1.86 \times 1 = 3.72 \ K$.
The freezing point of pure water is $0 \ ^oC$ $(273.15 \ K)$.
Freezing point of solution = $T_f^o - \Delta T_f = 0 \ ^oC - 3.72 \ ^oC = -3.72 \ ^oC$.

(A) For $KCl$,the van't Hoff factor $i = 2$ (as $KCl \rightarrow K^+ + Cl^-$).
For $BaCl_2$,the van't Hoff factor $i = 3$ (as $BaCl_2 \rightarrow Ba^{2+} + 2Cl^-$).
Since the concentration is the same,the depression in freezing point $\Delta T_f$ is directly proportional to $i$ $(\Delta T_f = i \times K_f \times m)$.
Therefore,$\frac{\Delta T_f(KCl)}{\Delta T_f(BaCl_2)} = \frac{i(KCl)}{i(BaCl_2)} = \frac{2}{3}$.
Given $\Delta T_f(KCl) = 0 - (-2) = 2 \ ^\circ C$.
So,$\frac{2}{\Delta T_f(BaCl_2)} = \frac{2}{3}$,which gives $\Delta T_f(BaCl_2) = 3 \ ^\circ C$.
The freezing point of the $BaCl_2$ solution is $0 - 3 = -3 \ ^\circ C$.

(A) The degree of ionization $\alpha = 0.20$ and molality $m = 0.2 \ m$.
For the dissociation of $HX \rightleftharpoons H^+ + X^-$,the van't Hoff factor $i = 1 + \alpha = 1 + 0.2 = 1.2$.
The depression in freezing point is given by $\Delta T_f = i \times K_f \times m$.
$\Delta T_f = 1.2 \times 1.86 \times 0.2 = 0.4464 \ \approx 0.45 \ K$.
The freezing point of the solution is $T_f = T_f^0 - \Delta T_f = 0 - 0.45 = -0.45 \ ^oC$.

(C) Assuming $100\%$ ionization:
For $K_3[Fe(CN)_6]$,the van't Hoff factor $i = 3 + 1 = 4$.
For $Al(NO_3)_3$,the dissociation is $Al(NO_3)_3 \rightarrow Al^{3+} + 3NO_3^-$,so $i = 1 + 3 = 4$.
Since both have $i = 4$,$Al(NO_3)_3$ has the same van't Hoff factor as $K_3[Fe(CN)_6]$.

(D) The depression in freezing point $(\Delta T_f)$ is a colligative property,which depends on the van't Hoff factor $(i)$.
$\Delta T_f = i \times K_f \times m$.
For a given concentration $(m = 0.1 \, M)$,$\Delta T_f$ is directly proportional to $i$.
$1$. For $NaCl$,$i = 2$ $(Na^+ + Cl^-)$.
$2$. For Urea,$i = 1$ (non-electrolyte).
$3$. For Glucose,$i = 1$ (non-electrolyte).
$4$. For $K_2SO_4$,$i = 3$ $(2K^+ + SO_4^{2-})$.
Since $K_2SO_4$ has the highest van't Hoff factor $(i = 3)$,it will show the maximum depression in freezing point.