Colligative properties of electrolyte Questions in English

(B) The dissociation of $2-$iodopropanoic acid $(CH_3CH(I)COOH)$ is given by: $CH_3CH(I)COOH \rightleftharpoons CH_3CH(I)COO^- + H^+$.
For a weak acid,the van't Hoff factor $i = 1 + \alpha(n-1)$.
Here,$n = 2$ and $\alpha = 0.05$ $(5 \%)$.
So,$i = 1 + 0.05(2-1) = 1.05$.
The depression in freezing point is given by $\Delta T_f = i \times K_f \times m$.
Given $K_f = 1.86 \, K \, kg \, mol^{-1}$ and $m = 0.1 \, molal$.
$\Delta T_f = 1.05 \times 1.86 \times 0.1 = 0.1953 \, K$.
The freezing point of the solution is $T_f = T_f^0 - \Delta T_f = 0 - 0.1953 = -0.1953 \, ^oC$.

(A) For an electrolyte $A_xB_y$ dissociating as $A_xB_y \rightarrow xA^{y+} + yB^{x-}$,the total number of ions produced is $n = x + y$.
If '$\alpha$' is the degree of dissociation,the number of particles at equilibrium is $1 - \alpha + x\alpha + y\alpha = 1 + \alpha(x + y - 1)$.
The van't Hoff factor '$i$' is defined as the ratio of the number of particles after dissociation to the number of particles before dissociation.
Thus,$i = 1 + \alpha(x + y - 1)$.
Rearranging for '$\alpha$',we get $\alpha = \frac{i - 1}{x + y - 1}$.

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

(B) The depression in freezing point is given by $\Delta T_f = T_f^\circ - T_f = 0 - (-0.063) = 0.063 \ K$.
Using the formula $\Delta T_f = i \times K_f \times m$,we have $0.063 = i \times 1.86 \times 0.011$.
Calculating $i$: $i = \frac{0.063}{1.86 \times 0.011} = \frac{0.063}{0.02046} \approx 3.079$.
For $K_3[Fe(CN)_6]$,the dissociation is $K_3[Fe(CN)_6] \rightarrow 3K^+ + [Fe(CN)_6]^{3-}$,so $n = 4$.
The degree of dissociation $\alpha$ is given by $\alpha = \frac{i - 1}{n - 1}$.
Substituting the values: $\alpha = \frac{3.079 - 1}{4 - 1} = \frac{2.079}{3} = 0.693$.
Percentage of dissociation $\approx 69.3 \%$,which is closest to $67 \%$.

(B) The reaction is $2K^+ + 2CN^- + Hg(CN)_2 \to 2K^+ + [Hg(CN)_4]^{2-}$.
Initially,there are $2$ moles of $K^+$ and $2$ moles of $CN^-$ ions,totaling $4$ particles per formula unit of $KCN$ (assuming complete dissociation).
After the reaction,the $CN^-$ ions are consumed to form the complex $[Hg(CN)_4]^{2-}$.
The total number of particles in the solution decreases because $4$ ions $(2K^+ + 2CN^-)$ are replaced by $3$ particles $(2K^+ + [Hg(CN)_4]^{2-})$.
Since colligative properties like osmotic pressure are directly proportional to the number of particles in the solution,a decrease in the number of particles leads to a decrease in osmotic pressure.

(B) The vapor pressure of a solution depends on the van't Hoff factor $(i)$ and the molality $(m)$ of the solution. For the vapor pressures to be equal,the effective concentration (osmolarity) must be equal: $i_1 m_1 = i_2 m_2 = i_3 m_3 = i_4 m_4$.
For $KCl$,$i = 2$ $(K^+ + Cl^-)$,so $i \times m = 2 \times 1.2 = 2.4$.
For $Na_2SO_4$,$i = 3$ $(2Na^+ + SO_4^{2-})$,so $3 \times m_1 = 2.4 \implies m_1 = 0.8 \, m$.
For urea (non-electrolyte),$i = 1$,so $1 \times m_2 = 2.4 \implies m_2 = 2.4 \, m$.
For $AlCl_3$,$i = 4$ $(Al^{3+} + 3Cl^-)$,so $4 \times m_3 = 2.4 \implies m_3 = 0.6 \, m$.
Thus,the molalities are $0.8 \, m, 2.4 \, m, 0.6 \, m$.

(D) The elevation in boiling point is given by the formula $\Delta T_b = i \times K_b \times m$.
Since $1 \ mol$ of each compound is dissolved in $1 \ L$ of solution,the molality $(m)$ is the same for all.
Therefore,$\Delta T_b$ is directly proportional to the van't Hoff factor $(i)$.
The van't Hoff factor $(i)$ depends on the degree of dissociation $(\alpha)$.
Stronger acids dissociate more completely in water. The acid strength order for hydrogen halides is $HI > HBr > HCl > HF$.
This is because the bond dissociation energy follows the order $HI < HBr < HCl < HF$.
Since $HI$ has the lowest bond dissociation energy,it dissociates most readily,leading to the highest number of ions in the solution.
Thus,$i$ is maximum for $HI$,resulting in the largest $\Delta T_b$ value.

(A) The depression in freezing point is given by $\Delta T_f = i \times K_f \times m$.
Since the concentration $(m)$ and solvent $(K_f)$ are the same for both solutions,$\Delta T_f \propto i$.
For $KCl$,$i = 2$ $(KCl \rightarrow K^+ + Cl^-)$.
For $BaCl_2$,$i = 3$ $(BaCl_2 \rightarrow Ba^{2+} + 2Cl^-)$.
Given $\Delta T_f$ for $KCl = 0 - (-2) = 2 \ ^oC$.
Using the ratio: $\frac{\Delta T_f(KCl)}{\Delta T_f(BaCl_2)} = \frac{i(KCl)}{i(BaCl_2)} = \frac{2}{3}$.
$\Delta T_f(BaCl_2) = \frac{3 \times 2}{2} = 3 \ ^oC$.
Freezing point of $BaCl_2 = 0 - 3 = -3 \ ^oC$.

(D) The depression in freezing point is given by the formula $\Delta T_f = i \times K_f \times m$.
For equimolal solutions,$\Delta T_f$ is directly proportional to the van't Hoff factor $(i)$.
$C_6H_{12}O_6$ (glucose) is a non-electrolyte,so its $i = 1$.
$C_6H_5NH_3^+Cl^-$ 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 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 $C_6H_{12}O_6$.

Number of moles of acetic acid = $\frac{0.6 \, mL \times 1.06 \, g \, mL^{-1}}{60 \, g \, mol^{-1}} = 0.0106 \, mol$.
Molality $(m) = \frac{0.0106 \, mol}{1 \, kg} = 0.0106 \, mol \, kg^{-1}$.
Theoretical depression in freezing point $(\Delta T_{f, \text{calc}}) = K_{f} \times m = 1.86 \, K \, kg \, mol^{-1} \times 0.0106 \, mol \, kg^{-1} = 0.0197 \, K$.
van't Hoff Factor $(i) = \frac{\Delta T_{f, \text{obs}}}{\Delta T_{f, \text{calc}}} = \frac{0.0205 \, K}{0.0197 \, K} \approx 1.041$.
For dissociation $CH_{3}COOH \rightleftharpoons CH_{3}COO^{-} + H^{+}$,$i = 1 + \alpha$,where $\alpha$ is the degree of dissociation.
$\alpha = i - 1 = 1.041 - 1 = 0.041$.
Dissociation constant $(K_{a}) = \frac{c \alpha^{2}}{1 - \alpha} = \frac{0.0106 \times (0.041)^{2}}{1 - 0.041} = \frac{0.0106 \times 0.001681}{0.959} \approx 1.86 \times 10^{-5}$.

(D) Molar mass of $CH_3CH_2CHClCOOH = (3 \times 12) + (5 \times 1) + 35.5 + (2 \times 16) + 1 = 122.5 \ g \ mol^{-1}$.
Number of moles of $CH_3CH_2CHClCOOH = \frac{10 \ g}{122.5 \ g \ mol^{-1}} = 0.0816 \ mol$.
Molality $(m)$ of the solution $= \frac{0.0816 \ mol}{0.250 \ kg} = 0.3264 \ mol \ kg^{-1}$.
For the dissociation $CH_3CH_2CHClCOOH \leftrightarrow CH_3CH_2CHClCOO^{-} + H^{+}$,the dissociation constant $K_a$ is given by $K_a = \frac{C \alpha^2}{1 - \alpha} \approx C \alpha^2$.
$\alpha = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{1.4 \times 10^{-3}}{0.3264}} = 0.0655$.
The van't Hoff factor $i = 1 + \alpha = 1 + 0.0655 = 1.0655$.
The depression in freezing point $\Delta T_f = i \cdot K_f \cdot m = 1.0655 \times 1.86 \ K \ kg \ mol^{-1} \times 0.3264 \ mol \ kg^{-1} = 0.647 \ K \approx 0.65 \ K$.

(N/A) It is given that:
$w_1 = 500 \ g$
$w_2 = 19.5 \ g$
$K_f = 1.86 \ K \ kg \ mol^{-1}$
$\Delta T_f = 1 \ K$
We know that:
$M_2 = \frac{K_f \times w_2 \times 1000}{\Delta T_f \times w_1}$
$= \frac{1.86 \ K \ kg \ mol^{-1} \times 19.5 \ g \times 1000 \ g \ kg^{-1}}{500 \ g \times 1 \ K}$
$= 72.54 \ g \ mol^{-1}$
Therefore,observed molar mass of $CH_2FCOOH$,$(M_2)_{obs} = 72.54 \ g \ mol^{-1}$
The calculated molar mass of $CH_2FCOOH$ is $(M_2)_{cal} = 12 + 2 + 19 + 12 + 16 + 16 + 1 = 78 \ g \ mol^{-1}$
Therefore,van't Hoff factor,$i = \frac{(M_2)_{cal}}{(M_2)_{obs}} = \frac{78}{72.54} = 1.0753$
For the dissociation reaction: $CH_2FCOOH \rightleftharpoons CH_2FCOO^{-} + H^{+}$
At equilibrium: $C(1-\alpha)$,$C\alpha$,$C\alpha$
Total moles $= C(1+\alpha)$
$\therefore i = 1 + \alpha \Rightarrow \alpha = i - 1 = 0.0753$
Concentration $C = \frac{19.5 \ g}{78 \ g \ mol^{-1}} \times \frac{1000 \ g}{500 \ g} \approx 0.5 \ M$
$K_a = \frac{C\alpha^2}{1 - \alpha} = \frac{0.5 \times (0.0753)^2}{1 - 0.0753} = \frac{0.5 \times 0.00567}{0.9247} \approx 3.07 \times 10^{-3}$

(0.0558 K) The dissociation of sodium sulphate is given by: $Na_2SO_4 \rightarrow 2Na^{+} + SO_4^{2-}$.
Since complete ionization occurs,the van't Hoff factor $(i)$ is $3$.
The molality $(m)$ of the solution is $0.01 \ mol / 1 \ kg = 0.01 \ m$.
The depression in freezing point $(\Delta T_f)$ is calculated using the formula: $\Delta T_f = i \times K_f \times m$.
Substituting the values: $\Delta T_f = 3 \times 1.86 \ K \ kg \ mol^{-1} \times 0.01 \ m = 0.0558 \ K$.

(N/A) The van't Hoff factor,denoted by $i$,was introduced to account for the extent of dissociation or association of solute particles in a solution. It is defined as:
$i = \frac{\text{Normal molar mass}}{\text{Abnormal molar mass}} = \frac{\text{Observed colligative property}}{\text{Calculated colligative property}}$
Alternatively,it can be expressed as:
$i = \frac{\text{Total number of moles of particles after association/dissociation}}{\text{Number of moles of particles before association/dissociation}}$
Here,the abnormal molar mass is the experimentally determined molar mass,and the calculated colligative properties are obtained by assuming that the non-volatile solute is neither associated nor dissociated.
In the case of association,the value of $i$ is less than unity $(i < 1)$,while for dissociation,it is greater than unity $(i > 1)$. When there is no association or dissociation,$i = 1$.
For example,the value of $i$ for an aqueous $KCl$ solution is close to $2$,while the value for ethanoic acid in benzene is nearly $0.5$.
Inclusion of the van't Hoff factor modifies the equations for colligative properties as follows:
$1$. Relative lowering of vapour pressure: $\frac{P_1^0 - P_1}{P_1^0} = i \frac{n_2}{n_1}$
$2$. Elevation of boiling point: $\Delta T_b = i K_b m$
$3$. Depression of freezing point: $\Delta T_f = i K_f m$
$4$. Osmotic pressure: $\Pi = i CRT$ (where $C = n/V$)

(B) The van't Hoff factor $(i)$ represents the number of particles a solute dissociates into in a solution.
For $K_{2}SO_{4}$,the dissociation is: $K_{2}SO_{4} \rightarrow 2K^{+} + SO_{4}^{2-}$. The total number of ions produced is $2 + 1 = 3$. Thus,$i = 3$.
For $K_{4}[Fe(CN)_{6}]$,the dissociation is: $K_{4}[Fe(CN)_{6}] \rightarrow 4K^{+} + [Fe(CN)_{6}]^{4-}$. The total number of ions produced is $4 + 1 = 5$. Thus,$i = 5$.

(B) The elevation in boiling point is given by $\Delta T_{b} = i \times K_{b} \times m$.
For $0.05 \ m$ $CaCl_{2}$ solution,$i = 3$ (since $CaCl_{2} \rightarrow Ca^{2+} + 2Cl^{-}$). Thus,$\Delta T_{b(CaCl_{2})} = 3 \times 0.05 \times K_{b} = 0.15 \times K_{b}$.
For $0.10 \ m$ $CrCl_{3} \cdot xNH_{3}$ solution,$\Delta T_{b(complex)} = i \times 0.10 \times K_{b}$.
Given $\Delta T_{b(complex)} = 2 \times \Delta T_{b(CaCl_{2})}$,we have $i \times 0.10 \times K_{b} = 2 \times (0.15 \times K_{b}) = 0.30 \times K_{b}$.
Therefore,$i = 3$.
Since the complex $[Cr(NH_{3})_{x}Cl_{3}]$ ionizes to give $i = 3$ ions,it must dissociate as $[Cr(NH_{3})_{x}Cl_{3-y}]^{y+} + yCl^{-}$,where $1 + y = 3$,so $y = 2$.
This means the complex is $[Cr(NH_{3})_{x}Cl]Cl_{2}$.
Given the coordination number of $Cr$ is $6$,the number of ligands is $x + 1 = 6$,which gives $x = 5$.

(D) For the dissociation reaction: $AB_2 \rightleftharpoons A^{2+} + 2B^-$,the number of ions produced per formula unit is $n = 3$.
Given the degree of dissociation $\alpha = 10\% = 0.1$.
The van't Hoff factor $i$ is calculated as $i = 1 + (n-1)\alpha = 1 + (3-1)0.1 = 1 + 0.2 = 1.2$.
The elevation in boiling point is given by $\Delta T_b = i \cdot K_b \cdot m$.
Substituting the values: $\Delta T_b = 1.2 \times 0.5 \text{ K kg mol}^{-1} \times 10.0 \text{ mol kg}^{-1} = 6 \text{ K}$ (or $6^\circ C$).
The boiling point of the solution is $T_b = T_b^\circ + \Delta T_b = 100^\circ C + 6^\circ C = 106^\circ C$.

(C) Molar mass of $ClCH_2COOH = 94.5 \text{ g mol}^{-1}$.
Molality $(m) = \frac{9.45 \text{ g}}{94.5 \text{ g mol}^{-1} \times 0.5 \text{ kg}} = 0.2 \text{ m}$.
Using the formula $\Delta T_f = i \cdot K_f \cdot m$:
$0.5 = i \times 1.86 \times 0.2$.
$i = \frac{0.5}{0.372} \approx 1.344$.
For the dissociation $ClCH_2COOH \rightleftharpoons ClCH_2COO^- + H^+$,the van't Hoff factor is $i = 1 + \alpha$.
$\alpha = i - 1 = 0.344$.
The dissociation constant $K_a = \frac{C\alpha^2}{1-\alpha} = \frac{0.2 \times (0.344)^2}{1 - 0.344} = \frac{0.2 \times 0.118336}{0.656} \approx 0.0361 = 36.1 \times 10^{-3}$.
Thus,$x \approx 36$.

(C) The electrolyte $A_{2}B_{3}$ dissociates as: $A_{2}B_{3} \rightarrow 2A^{3+} + 3B^{2-}$.
The number of ions produced per formula unit is $n = 2 + 3 = 5$.
The degree of dissociation $\alpha = 0.60$.
The van't Hoff factor $i = 1 + \alpha(n - 1) = 1 + 0.60(5 - 1) = 1 + 0.60(4) = 1 + 2.4 = 3.4$.
The elevation in boiling point is given by $\Delta T_{b} = i K_{b} m$.
Substituting the values: $\Delta T_{b} = 3.4 \times 0.52 \times 1 = 1.768 \ K$.
The boiling point of the solution $T_{b} = T_{b}^{\circ} + \Delta T_{b} = 373.15 + 1.768 = 374.918 \ K$.
Rounding off to the nearest integer,we get $375 \ K$.

(B) The degree of dissociation $\alpha = 0.75$ and for $AB$,the number of ions $n = 2$.
The van't Hoff factor $i = 1 - \alpha + n\alpha = 1 - 0.75 + 2 \times 0.75 = 1.75$.
The elevation in boiling point is given by $\Delta T_{b} = i \times K_{b} \times m$.
Substituting the values: $2.5 = 1.75 \times 0.52 \times m$.
Calculating molality $m = \frac{2.5}{1.75 \times 0.52} = \frac{2.5}{0.91} \approx 2.747$.
Rounding off to the nearest integer,we get $3$.

(A) The freezing point depression is given by $\Delta T_f = i \times K_f \times m$,where $i$ is the van't Hoff factor and $m$ is the molality.
Since $0.10 \, M \, C_{2}H_{5}OH$ is a non-electrolyte,its van't Hoff factor $i = 1$. Thus,its effective concentration is $1 \times 0.10 = 0.10 \, M$.
For the other solutions:
$(i)$ $Ba_{3}(PO_{4})_{2}$ dissociates into $5$ ions $(3Ba^{2+} + 2PO_{4}^{3-})$,so $i = 5$. Effective concentration = $5 \times 0.10 = 0.50 \, M$.
$(ii)$ $Na_{2}SO_{4}$ dissociates into $3$ ions $(2Na^{+} + SO_{4}^{2-})$,so $i = 3$. Effective concentration = $3 \times 0.10 = 0.30 \, M$.
$(iii)$ $KCl$ dissociates into $2$ ions $(K^{+} + Cl^{-})$,so $i = 2$. Effective concentration = $2 \times 0.10 = 0.20 \, M$.
$(iv)$ $Li_{3}PO_{4}$ dissociates into $4$ ions $(3Li^{+} + PO_{4}^{3-})$,so $i = 4$. Effective concentration = $4 \times 0.10 = 0.40 \, M$.
Since all these solutions have an effective concentration greater than $0.10 \, M$,they all exhibit a greater depression in freezing point,meaning their freezing points are lower than that of $0.10 \, M \, C_{2}H_{5}OH$.
Therefore,the total number of such solutions is $4$.

(D) The freezing point depression $(\Delta T_{f})$ is a colligative property given by the formula $\Delta T_{f} = i \times K_{f} \times m$,where $i$ is the Van't Hoff factor.
For a given molality $(m)$ and solvent,$\Delta T_{f}$ is directly proportional to the Van't Hoff factor $(i)$.
$1.$ Hydrazine $(NH_{2}NH_{2})$,Glucose $(C_{6}H_{12}O_{6})$,and Glycine $(NH_{2}CH_{2}COOH)$ are non-electrolytes,so their $i \approx 1$.
$2.$ $KHSO_{4}$ is a strong electrolyte that dissociates in water as $KHSO_{4} \rightarrow K^{+} + H^{+} + SO_{4}^{2-}$,giving $i \approx 3$.
Since $KHSO_{4}$ has the highest Van't Hoff factor,it will exhibit the largest freezing point depression.

(D) The freezing point depression is given by the formula $\Delta T_{f} = i \times K_{f} \times m$.
Since $K_{f}$ and $m$ are constant for all solutions,$\Delta T_{f} \propto i$.
The freezing point $(T_{f})$ is related to the depression as $T_{f} = T_{f}^{\circ} - \Delta T_{f}$,meaning $T_{f} \propto -i$.
Therefore,the solution with the highest van't Hoff factor $(i)$ will have the lowest freezing point.
For $C_{6}H_{12}O_{6}$ (non-electrolyte),$i = 1$.
For $K_{2}SO_{4} \rightarrow 2K^{+} + SO_{4}^{2-}$,$i = 3$.
For $KI \rightarrow K^{+} + I^{-}$,$i = 2$.
For $Al_{2}(SO_{4})_{3} \rightarrow 2Al^{3+} + 3SO_{4}^{2-}$,$i = 5$.
Since $Al_{2}(SO_{4})_{3}$ has the highest value of $i = 5$,it will have the lowest freezing point.

(C) The boiling point elevation is given by the formula $\Delta T_{b} = i \times K_{b} \times m$.
For $CuSO_{4} \cdot 5H_{2}O$,the van't Hoff factor $i = 2$ because it dissociates as $CuSO_{4} \rightarrow Cu^{2+} + SO_{4}^{2-}$.
Given $K_{b} = 0.512 \, K \, kg \, mol^{-1}$ and molality $m = 0.1 \, mol \, kg^{-1}$.
$\Delta T_{b} = 2 \times 0.512 \times 0.1 = 0.1024 \, K$.
The boiling point of pure water is $100^{\circ}C$.
Therefore,the boiling point of the solution is $T_{b} = 100 + 0.1024 = 100.1024^{\circ}C$,which is closest to $100.10^{\circ}C$.

(A) The elevation in boiling point 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,$\Delta T_{b} \propto i$ (van't Hoff factor).
$CH_{3}COOH$ is a weak electrolyte and dissociates partially,so its $i$ value is approximately $1 < i < 2$.
$NaCl$ dissociates into $2$ ions $(Na^+, Cl^-)$,so $i \approx 2$.
$Na_{2}SO_{4}$ dissociates into $3$ ions $(2Na^+, SO_{4}^{2-})$,so $i \approx 3$.
$K_{3}PO_{4}$ dissociates into $4$ ions $(3K^+, PO_{4}^{3-})$,so $i \approx 4$.
Therefore,the order of boiling points is $CH_{3}COOH < NaCl < Na_{2}SO_{4} < K_{3}PO_{4}$.

(D)
Boiling point elevation is a colligative property,which depends on the number of solute particles in the solution.
For the same molar concentration $(0.01 \, M)$,the solution with the highest van't Hoff factor $(i)$ will show the greatest elevation in boiling point.
$1$. Sucrose $(C_{12}H_{22}O_{11})$: Non-electrolyte,$i = 1$.
$2$. $NaCl$: Dissociates as $NaCl \rightarrow Na^{+} + Cl^{-}$,$i = 2$.
$3$. $CaCl_2$: Dissociates as $CaCl_2 \rightarrow Ca^{2+} + 2Cl^{-}$,$i = 3$.
Since $CaCl_2$ produces the maximum number of particles ($3$ ions per formula unit),it will have the highest boiling point.

(D) For isotonic solutions, the osmotic pressure $(\pi)$ is equal: $\pi_{K_2SO_4} = \pi_{\text{glucose}}$.
Since $\pi = iCRT$, we have $i_{K_2SO_4} \times 0.004 \times RT = 1 \times 0.01 \times RT$.
Solving for the van't Hoff factor $(i)$: $i = \frac{0.01}{0.004} = 2.5$.
For the dissociation of $K_2SO_4$ $(K_2SO_4 \rightarrow 2K^+ + SO_4^{2-})$, the number of ions produced $(n)$ is $3$.
The degree of dissociation $(\alpha)$ is given by the formula: $i = 1 + (n - 1)\alpha$.
Substituting the values: $2.5 = 1 + (3 - 1)\alpha$.
$1.5 = 2\alpha$, which gives $\alpha = 0.75$.
The percentage dissociation is $\alpha \times 100 = 75\%$.

(B) $MgCl_2 \rightarrow Mg^{2+} + 2Cl^-$
Initial moles: $1$,$0$,$0$
At equilibrium: $1-\alpha$,$\alpha$,$2\alpha$
Total particles $i = 1 + 2\alpha$. Given $\alpha = 0.8$,so $i = 1 + 2(0.8) = 2.6$.
For a $1.0$ molal solution,$n_2 = 1$ mole of solute in $1000 \ g$ of water ($n_1 = 1000/18 = 55.55$ moles).
Using Raoult's Law: $\frac{p^{\circ} - p_s}{p^{\circ}} = \frac{i \times n_2}{n_1 + i \times n_2} \approx \frac{i \times n_2}{n_1}$.
$\frac{50 - p_s}{50} = \frac{2.6 \times 1}{55.55} = 0.0468$.
$50 - p_s = 2.34 \implies p_s = 47.66 \ mm \ Hg$.
Rounding to the nearest integer,$p_s \approx 48 \ mm \ Hg$.

(A) The van 't Hoff factor $(i)$ for a strong electrolyte like $NaCl$ increases as the concentration decreases.
In more concentrated solutions (like $0.1 \ M$),interionic attractions are stronger,which reduces the effective dissociation.
As the solution becomes more dilute (from $A$ to $C$),the ions move further apart,and $i$ approaches its theoretical maximum value of $2$.
Therefore,the order of the van 't Hoff factor is $i_A < i_B < i_C$.

(A) Mass of $CH_3COOH = V \times d = 5 \ mL \times 1.2 \ g \ mL^{-1} = 6 \ g$.
Moles of $CH_3COOH = \frac{6 \ g}{60 \ g \ mol^{-1}} = 0.1 \ mol$.
Molality $(m) = \frac{0.1 \ mol}{1 \ kg \ \text{water}} = 0.1 \ m$.
For the dissociation $CH_3COOH \rightleftharpoons CH_3COO^{-} + H^{+}$,$K_{a} = \frac{C\alpha^{2}}{1-\alpha}$.
Since $\alpha$ is small,$1-\alpha \approx 1$,so $K_{a} \approx C\alpha^{2}$.
$\alpha = \sqrt{\frac{K_{a}}{C}} = \sqrt{\frac{6.25 \times 10^{-5}}{0.1}} = \sqrt{6.25 \times 10^{-4}} = 25 \times 10^{-3} = 0.025$.
van't Hoff factor $(i) = 1 + \alpha(n-1) = 1 + 0.025(2-1) = 1.025$.
Freezing point depression $\Delta T_{f} = i \times K_{f} \times m = 1.025 \times 1.86 \times 0.1 = 0.19065 \ {}^{\circ}C$.
Given $\Delta T_{f} = x \times 10^{-2} \ {}^{\circ}C$,so $x \approx 19$.

(A) The dissociation of $HX$ is given by: $HX \rightleftharpoons H^{+} + X^{-}$.
Initial concentration: $0.03 \ M$.
Equilibrium concentration: $(0.03 - x), x, x$.
Since $K_{a}$ is very small,$0.03 - x \approx 0.03$.
$K_{a} = \frac{x^2}{0.03} = 1.2 \times 10^{-5}$.
$x^2 = 3.6 \times 10^{-7} = 36 \times 10^{-8}$.
$x = 6 \times 10^{-4} \ M$.
Total concentration of particles $C_{total} = (0.03 - x) + x + x = 0.03 + x = 0.03 + 0.0006 = 0.0306 \ M$.
Osmotic pressure $\Pi = C_{total} \times R \times T$.
$\Pi = 0.0306 \times 0.083 \times 300 = 0.76194 \ bar$.
$\Pi = 76.194 \times 10^{-2} \ bar$.
Rounding to the nearest integer,we get $76 \times 10^{-2} \ bar$.

(B) The dissociation reaction is $AB_2 \rightarrow A^{2+} + 2B^-$. The van't Hoff factor $i$ is given by $i = 1 + (n-1)\alpha$,where $n=3$. So,$i = 1 + 2\alpha$.
The boiling point elevation is $\Delta T_b = T_b - T_b^{\circ} = 100.52^{\circ} C - 100^{\circ} C = 0.52 \ K$.
The molality $m$ is calculated as $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{10/200}{100/1000} = \frac{0.05}{0.1} = 0.5 \ mol \ kg^{-1}$.
Using the formula $\Delta T_b = i \cdot K_b \cdot m$:
$0.52 = (1 + 2\alpha) \times 0.52 \times 0.5$
$1 = (1 + 2\alpha) \times 0.5$
$2 = 1 + 2\alpha$
$2\alpha = 1$
$\alpha = 0.5 = 5 \times 10^{-1}$.
Thus,the value is $5$.

(A) The dissociation of $K_3[Fe(CN)_6]$ is given by: $K_3[Fe(CN)_6] \rightarrow 3K^{+} + [Fe(CN)_6]^{3-}$.
Since $1 \ mol$ of $K_3[Fe(CN)_6]$ produces $4 \ mol$ of ions,the van't Hoff factor $i = 4$.
The formula for depression in freezing point is $\Delta T_f = i \times K_f \times m$.
Here,molality $m = \frac{w_2 \times 1000}{M_2 \times w_1} = \frac{0.1 \times 1000}{329 \times 100} = \frac{1}{329} \ mol \ kg^{-1}$.
$\Delta T_f = 4 \times 1.86 \times \frac{1}{329} \approx 0.0226 \ K \approx 0.023 \ K$.
Since the freezing point of pure water is $0 \ {}^{\circ}C$,the freezing point of the solution is $T_f = 0 - \Delta T_f = -0.023 \ {}^{\circ}C = -2.3 \times 10^{-2} \ {}^{\circ}C$.

(B) The dissociation reaction is: $MX_{2} \rightleftharpoons M^{2+} + 2X^{-}$.
Initially,let the concentration be $C$. At equilibrium,the concentrations are: $[MX_{2}] = C(1-\alpha)$,$[M^{2+}] = C\alpha$,and $[X^{-}] = 2C\alpha$.
The total number of particles (van't Hoff factor $i$) is given by: $i = 1 - \alpha + n\alpha$,where $n$ is the number of ions produced per formula unit.
For $MX_{2}$,$n = 3$ ($1$ $M^{2+}$ and $2$ $X^{-}$ ions).
Given $\alpha = 0.5$,we have: $i = 1 + (3-1)\alpha = 1 + 2(0.5) = 1 + 1 = 2$.
The ratio of the observed depression of freezing point to the theoretical depression (in the absence of dissociation) is equal to the van't Hoff factor $i$.
Therefore,the ratio is $2$.

(A) The depression in freezing point is given by $\Delta T_{f} = i K_{f} m$.
Given $\Delta T_{f} = 0 - (-0.0558) = 0.0558 \ K$,$K_{f} = 1.86 \ K \ kg \ mol^{-1}$,and $m = 0.01 \ m$.
Substituting the values: $0.0558 = i \times 1.86 \times 0.01$.
$i = \frac{0.0558}{0.0186} = 3$.
Since the complex is a cobalt $(III)$ chloride-ammonia complex,it is of the form $[Co(NH_3)_xCl_y]Cl_z$.
For $i = 3$,the complex dissociates into $3$ ions: $[Co(NH_3)_xCl_y]^{2+} + 2Cl^-$.
This implies there are $2$ chloride ions outside the coordination sphere.
Since the coordination number of $Co(III)$ is $6$,the formula is $[Co(NH_3)_5Cl]Cl_2$.
Thus,the number of chloride ions inside the coordination sphere is $1$.

(A) Given: $n_{\text{salt}} = 0.1 \ mol$,$W_{\text{water}} = 1.8 \ kg = 1800 \ g$,$T = 35^{\circ} C$,$\alpha = 0.90$,$P_s = 59.724 \ mm \ Hg$,$P^{\circ} = 60.000 \ mm \ Hg$.
Moles of water $n_w = \frac{1800 \ g}{18 \ g/mol} = 100 \ mol$.
Let $x$ be the number of ions per formula unit. The salt dissociates as $AB_x \rightarrow A^{x+} + xB^-$. The van't Hoff factor $i = 1 + \alpha(x - 1) = 1 + 0.9(x - 1) = 0.1 + 0.9x$.
Using Raoult's Law: $\frac{P^{\circ} - P_s}{P_s} = \frac{i \cdot n_{\text{salt}}}{n_w}$.
$\frac{60.000 - 59.724}{59.724} = \frac{(0.1 + 0.9x) \cdot 0.1}{100}$.
$\frac{0.276}{59.724} = \frac{0.01 + 0.09x}{100}$.
$0.004621 \approx \frac{0.01 + 0.09x}{100} \Rightarrow 0.4621 = 0.01 + 0.09x$.
$0.4521 = 0.09x \Rightarrow x = \frac{0.4521}{0.09} \approx 5.02$.
Thus,the number of ions per formula unit is $5$.

(D) For the dissociation of $A_2 B$: $A_2 B \rightarrow 2 A^{+} + B^{2-}$.
Here,the number of ions produced per formula unit is $y = 3$.
The degree of ionisation is $\alpha = 30 \% = 0.3$.
The van't Hoff factor $(i)$ is calculated using the formula: $i = 1 + (y - 1)\alpha$.
Substituting the values: $i = 1 + (3 - 1)(0.3) = 1 + (2)(0.3) = 1 + 0.6 = 1.6$.
Expressing $1.6$ in the form $............ \times 10^{-1}$,we get $16 \times 10^{-1}$.
Thus,the correct option is $D$.

(D) The dissociation reaction for the salt is: $MX_3 \rightarrow M^{3+} + 3X^{-}$.
Here,the number of ions produced per formula unit is $n = 1 + 3 = 4$.
The relationship between the van't Hoff factor $(i)$ and the degree of dissociation $(\alpha)$ is given by: $i = 1 + (n - 1)\alpha$.
Substituting the given values: $2 = 1 + (4 - 1)\alpha$.
$2 = 1 + 3\alpha$.
$3\alpha = 1$.
$\alpha = \frac{1}{3} \approx 0.3333$.
Therefore,the percentage dissociation is $0.3333 \times 100 = 33.33\%$.
The nearest integer is $33\%$.

(C) The formula for depression in freezing point is $\Delta T_f = i \times K_f \times m$,where $i$ is the van't Hoff factor,$K_f$ is the cryoscopic constant,and $m$ is the molality.
For $KCl$,which is a strong electrolyte,$i = 2$ (dissociates into $K^+$ and $Cl^-$).
Given: $\Delta T_f = 2 \ K$,$K_f = 1.86 \ K \ kg \ mol^{-1}$,$W_{solvent} = 500 \ g = 0.5 \ kg$,$MW_{KCl} = 74.5 \ g \ mol^{-1}$.
Molality $m = \frac{W_{solute} / MW_{solute}}{W_{solvent} (kg)} = \frac{W_{KCl} / 74.5}{0.5}$.
Substituting the values: $2 = 2 \times 1.86 \times \frac{W_{KCl} / 74.5}{0.5}$.
$2 = 3.72 \times \frac{W_{KCl}}{37.25}$.
$W_{KCl} = \frac{2 \times 37.25}{3.72} \approx 20.0 \ g$.

(D) The elevation in boiling point is given by $\Delta T_{b} = i \times K_{b} \times m$.
Since the concentration $(m)$ is the same for all solutions,the elevation in boiling point is directly proportional to the van't Hoff factor $(i)$.
$(i)$ For $1 \%$ urea,$i = 1$ (non-electrolyte).
$(ii)$ For $1 \%$ sucrose,$i = 1$ (non-electrolyte).
$(iii)$ For $1 \%$ NaCl,$i = 2$ (dissociates into $Na^{+}$ and $Cl^{-}$).
$(iv)$ For $1 \% CaCl_{2}$,$i = 3$ (dissociates into $Ca^{2+}$ and $2Cl^{-}$).
Since $CaCl_{2}$ has the highest van't Hoff factor $(i = 3)$,it will show the maximum elevation in boiling point. Therefore,the $1 \% CaCl_{2}$ solution has the highest boiling point.

(A) Isotonic solutions have the same molar concentration of solute particles in the solution.
We calculate the molar concentration of particles (van't Hoff factor $i \times M$):
For $A$ (glucose): $1 \times 0.1 \ M = 0.1 \ M$.
For $B$ $(NaCl)$: $2 \times 0.05 \ M = 0.1 \ M$.
For $C$ $(BaCl_{2})$: $3 \times 0.05 \ M = 0.15 \ M$.
For $D$ $(AlCl_{3})$: $4 \times 0.1 \ M = 0.4 \ M$.
Since solutions $A$ and $B$ have the same concentration of particles $(0.1 \ M)$,they are isotonic.

(D) The elevation in boiling point is given by the formula $\Delta T_{b} = i \times K_{b} \times m$.
For a nonelectrolyte,the van't Hoff factor $i = 1$. Thus,for the nonelectrolyte solution,$\Delta T_{b} = 1 \times K_{b} \times m = x \ K$.
For $AlCl_3$,the dissociation is $AlCl_3 \rightarrow Al^{3+} + 3Cl^-$.
The van't Hoff factor $i$ for $AlCl_3$ is $1 + 3 = 4$.
Therefore,for $1 \ m \ AlCl_3$ solution,$\Delta T_{b} = 4 \times K_{b} \times m = 4 \times (K_{b} \times m) = 4x \ K$.

(D) The boiling point elevation is given by the formula $\Delta T_b = i \times K_b \times m$,where $i$ is the van't Hoff factor and $m$ is the molality. Since $K_b$ is constant,$\Delta T_b$ is proportional to $i \times m$.
For complete dissociation,$i$ equals the number of ions produced per formula unit.
$A$: $KCl \rightarrow K^+ + Cl^-$,$i = 2$. $\Delta T_b \propto 2 \times 0.2 = 0.4$.
$B$: $NaCl \rightarrow Na^+ + Cl^-$,$i = 2$. $\Delta T_b \propto 2 \times 0.1 = 0.2$.
$C$: $AlCl_3 \rightarrow Al^{3+} + 3Cl^-$,$i = 4$. $\Delta T_b \propto 4 \times 1 = 4.0$.
$D$: $MgCl_2 \rightarrow Mg^{2+} + 2Cl^-$,$i = 3$. $\Delta T_b \propto 3 \times 0.05 = 0.15$.
Comparing the values,$0.15 < 0.2 < 0.4 < 4.0$. Thus,$0.05 \ m \ MgCl_2$ exhibits the minimum boiling point elevation.

(C) The depression in freezing point is given by the formula $\Delta T_{f} = i \times K_{f} \times m$.
For a $1 \ m$ urea solution,urea is a non-electrolyte,so the van't Hoff factor $i = 1$. Thus,$\Delta T_{f} = 1 \times K_{f} \times 1 = K_{f} = x \ K$.
For a $1 \ m$ $CaCl_{2}$ solution,$CaCl_{2}$ dissociates as $CaCl_{2} \rightarrow Ca^{2+} + 2Cl^{-}$,so the van't Hoff factor $i = 3$.
Thus,$\Delta T_{f} = 3 \times K_{f} \times 1 = 3 \times x \ K = 3x \ K$.

(C) The elevation in boiling point 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,the elevation in boiling point depends directly on the van't Hoff factor $(i)$.
For complete dissociation:
$KCl \rightarrow K^+ + Cl^-$ $(i = 2)$
$NaCl \rightarrow Na^+ + Cl^-$ $(i = 2)$
$AlCl_3 \rightarrow Al^{3+} + 3Cl^-$ $(i = 4)$
$BaCl_2 \rightarrow Ba^{2+} + 2Cl^-$ $(i = 3)$
Since $AlCl_3$ has the highest van't Hoff factor $(i = 4)$,it will exhibit the maximum boiling point elevation.

(B) The depression in freezing point is given by the formula: $\Delta T_{f} = i \cdot K_{f} \cdot m$.
Given: $\Delta T_{f} = 0.660 \ K$,$m = 0.2 \ m$,and $K_{f} = 1.84 \ K \ kg \ mol^{-1}$.
Substituting the values: $0.660 = i \times 1.84 \times 0.2$.
$i = \frac{0.660}{1.84 \times 0.2} = \frac{0.660}{0.368} \approx 1.793$.
Thus,the van't Hoff factor $(i)$ is approximately $1.79$.

(D) For sucrose solution,$(\Delta T_f)_1 = K_f m_1 \dots (I)$.
For $CaCl_2$ solution,$(\Delta T_f)_2 = i K_f m_2 \dots (II)$.
Since $m_1 = m_2 = 1.25 \ m$,we have $\frac{(\Delta T_f)_2}{(\Delta T_f)_1} = \frac{i K_f m_2}{K_f m_1} = i$.
For $CaCl_2$,the van't Hoff factor $i = 3$ (as $CaCl_2 \rightarrow Ca^{2+} + 2Cl^-$).
Therefore,$(\Delta T_f)_2 = i \times (\Delta T_f)_1 = 3 \times x \ K = 3 x \ K$.

(A) The formula for depression in freezing point is $\Delta T_f = i \cdot K_f \cdot m$.
Given: $\Delta T_f = 0 - (-0.43) = 0.43 \ K$,$K_f = 1.86 \ K \ kg \ mol^{-1}$,and $m = 0.1 \ m$.
Substituting the values: $0.43 = i \times 1.86 \times 0.1$.
Solving for $i$: $i = \frac{0.43}{0.186} \approx 2.31$.
Rounding to the nearest given option,the value is $2.3$.

(D) Colligative properties depend on the number of particles in the solution. The relationship is given by the van't Hoff factor $(i)$.
For non-electrolytes like urea,glucose,and sucrose,the value of $i$ is $1$.
For sodium chloride $(NaCl)$,which is a strong electrolyte,it dissociates as $NaCl \rightarrow Na^+ + Cl^-$,resulting in $i = 2$.
Since the colligative property is directly proportional to the van't Hoff factor $(i)$,the solute with the highest $i$ value will exhibit the highest colligative property.