Question

At a given temperature the osmotic pressure of 0.1 M $Fe_2(SO_4)_3$ aqueous solution is nearest to that of 10L aqueous solution of: (Consider 100% dissociation of electrolytes) (If atomic mass of Ca = 40u, Cl = 35.5u, P = 31u, O = 15u, K = 39u, H = lu, N = 14u, Fe = 56u)

• A. 111 grams $CaCl_2$
• B. 310 grams $Ca_3[PO_4]_2$
• C. 342 grams $C_{12}H_{22}O_{11}$
• D. 368 grams $K_4[Fe(CN)_6]$

Answer 1

Answer: (D)

D

Answer 2

The osmotic pressure ($\Pi$) of a solution is given by the formula: $\Pi = iCRT$

where:

• $i$ is the van't Hoff factor (number of particles produced per formula unit of solute).
• $C$ is the molar concentration of the solute.
• $R$ is the ideal gas constant.
• $T$ is the temperature in Kelvin.

Since the temperature (T) and the gas constant (R) are the same for all solutions, the osmotic pressure will be proportional to the product $iC$. We need to find the option whose $iC$ value is closest to that of the given $Fe_2(SO_4)_3$ solution.

1. Calculate $iC$ for the reference solution: 0.1 M $Fe_2(SO_4)_3$

$Fe_2(SO_4)_3$ dissociates as: $Fe_2(SO_4)_3 \rightarrow 2Fe^{3+} + 3SO_4^{2-}$

Since 100% dissociation is assumed, the van't Hoff factor $i = 2 + 3 = 5$.

The concentration $C = 0.1$ M.

So, for $Fe_2(SO_4)_3$, $iC = 5 \times 0.1 = 0.5$.

2. Calculate $iC$ for each option:

Option (A): 111 grams $CaCl_2$ in 10L aqueous solution

• Molar mass of $CaCl_2$: $40 \text{ (Ca)} + 2 \times 35.5 \text{ (Cl)} = 40 + 71 = 111$ g/mol.
• Moles of $CaCl_2$: $\frac{111 \text{ g}}{111 \text{ g/mol}} = 1$ mol.
• Concentration (C): $\frac{1 \text{ mol}}{10 \text{ L}} = 0.1$ M.
• Dissociation: $CaCl_2 \rightarrow Ca^{2+} + 2Cl^-$
• Van't Hoff factor (i): $1 + 2 = 3$.
• $iC$ value: $3 \times 0.1 = 0.3$.
• Difference from target: $|0.5 - 0.3| = 0.2$.

Option (B): 310 grams $Ca_3[PO_4]_2$ in 10L aqueous solution

• Molar mass of $Ca_3[PO_4]_2$: $3 \times 40 \text{ (Ca)} + 2 \times (31 \text{ (P)} + 4 \times 15 \text{ (O)}) = 120 + 2 \times (31 + 60) = 120 + 2 \times 91 = 120 + 182 = 302$ g/mol.
• Moles of $Ca_3[PO_4]_2$: $\frac{310 \text{ g}}{302 \text{ g/mol}} \approx 1.0265$ mol.
• Concentration (C): $\frac{1.0265 \text{ mol}}{10 \text{ L}} = 0.10265$ M.
• Dissociation: $Ca_3[PO_4]_2 \rightarrow 3Ca^{2+} + 2PO_4^{3-}$
• Van't Hoff factor (i): $3 + 2 = 5$.
• $iC$ value: $5 \times 0.10265 = 0.51325$.
• Difference from target: $|0.5 - 0.51325| = 0.01325$.

Option (C): 342 grams $C_{12}H_{22}O_{11}$ in 10L aqueous solution

• Molar mass of $C_{12}H_{22}O_{11}$: $12 \times 12 \text{ (C)} + 22 \times 1 \text{ (H)} + 11 \times 15 \text{ (O)} = 144 + 22 + 165 = 331$ g/mol.
• Moles of $C_{12}H_{22}O_{11}$: $\frac{342 \text{ g}}{331 \text{ g/mol}} \approx 1.0332$ mol.
• Concentration (C): $\frac{1.0332 \text{ mol}}{10 \text{ L}} = 0.10332$ M.
• Dissociation: Sucrose is a non-electrolyte, so it does not dissociate.
• Van't Hoff factor (i): $1$.
• $iC$ value: $1 \times 0.10332 = 0.10332$.
• Difference from target: $|0.5 - 0.10332| = 0.39668$.

Option (D): 368 grams $K_4[Fe(CN)_6]$ in 10L aqueous solution

• Molar mass of $K_4[Fe(CN)_6]$: $4 \times 39 \text{ (K)} + 56 \text{ (Fe)} + 6 \times 12 \text{ (C)} + 6 \times 14 \text{ (N)} = 156 + 56 + 72 + 84 = 368$ g/mol.
• Moles of $K_4[Fe(CN)_6]$: $\frac{368 \text{ g}}{368 \text{ g/mol}} = 1$ mol.
• Concentration (C): $\frac{1 \text{ mol}}{10 \text{ L}} = 0.1$ M.
• Dissociation: $K_4[Fe(CN)_6] \rightarrow 4K^+ + [Fe(CN)_6]^{4-}$
• Van't Hoff factor (i): $4 + 1 = 5$.
• $iC$ value: $5 \times 0.1 = 0.5$.
• Difference from target: $|0.5 - 0.5| = 0$.

3. Comparison of $iC$ values:

• Reference $Fe_2(SO_4)_3$: $iC = 0.5$
• Option (A) $CaCl_2$: $iC = 0.3$ (Difference = 0.2)
• Option (B) $Ca_3[PO_4]_2$: $iC = 0.51325$ (Difference = 0.01325)
• Option (C) $C_{12}H_{22}O_{11}$: $iC = 0.10332$ (Difference = 0.39668)
• Option (D) $K_4[Fe(CN)_6]$: $iC = 0.5$ (Difference = 0)

Option (D) has an $iC$ value exactly equal to the reference solution, meaning its osmotic pressure is identical. Therefore, it is the nearest.