To compare osmotic pressures, use

$\pi = iMRT$

At the same temperature, solutions have the same osmotic pressure if they have the same value of $iM$.

So we compute the van’t Hoff factor $i$ for each solution assuming 100% ionization.

1. Compute $iM$ for each solution

A

1. $0.500\,\text{M } \mathrm{C_2H_5OH}$

   • Ethanol is a non-electrolyte, so $i=1$.
   • Hence, $iM = 1 \times 0.500 = 0.500$

2. $0.25\,\text{M } \mathrm{KBr}$

   • $\mathrm{KBr \to K^+ + Br^-}$, so $i=2$.
   • Hence, $iM = 2 \times 0.25 = 0.50$

So, pair A has same osmotic pressure.

B

1. $0.100\,\text{M } \mathrm{K_4[Fe(CN)_6]}$

   • $\mathrm{K_4[Fe(CN)_6] \to 4K^+ + [Fe(CN)_6]^{4-}}$
   • Total particles $=5$, so $i=5$.
   • Hence, $iM = 5 \times 0.100 = 0.50$

2. $0.100\,\text{M } \mathrm{FeSO_4(NH_4)_2SO_4}$

   • This is Mohr’s salt: $\mathrm{(NH_4)_2Fe(SO_4)_2}$
   • On ionization: $\mathrm{(NH_4)_2Fe(SO_4)_2 \to 2NH_4^+ + Fe^{2+} + 2SO_4^{2-}}$
   • Total particles $=5$, so $i=5$.
   • Hence, $iM = 5 \times 0.100 = 0.50$

So, pair B has same osmotic pressure.

C

1. $0.05\,\text{M } \mathrm{K_4[Fe(CN)_6]}$

   • $i=5$
   • Hence, $iM = 5 \times 0.05 = 0.25$

2. $0.25\,\text{M } \mathrm{NaCl}$

   • $\mathrm{NaCl \to Na^+ + Cl^-}$, so $i=2$.
   • Hence, $iM = 2 \times 0.25 = 0.50$

Since $0.25 \ne 0.50$, pair C does not have same osmotic pressure.

D

1. $0.15\,\text{M } \mathrm{NaCl}$

   • $i=2$
   • Hence, $iM = 2 \times 0.15 = 0.30$

2. $0.1\,\text{M } \mathrm{BaCl_2}$

   • $\mathrm{BaCl_2 \to Ba^{2+} + 2Cl^-}$, so $i=3$.
   • Hence, $iM = 3 \times 0.1 = 0.30$

So, pair D has same osmotic pressure.

E

1. $0.02\,\text{M } \mathrm{KCl\cdot MgCl_2\cdot 6H_2O}$
   • Water of crystallization does not contribute ions.
   • On dissociation: $\mathrm{KCl\cdot MgCl_2 \to K^+ + Mg^{2+} + 3Cl^-}$
   • Total particles $=5?$

Let us count carefully:

• from $\mathrm{KCl}$: $\mathrm{K^+ + Cl^-}$
• from $\mathrm{MgCl_2}$: $\mathrm{Mg^{2+} + 2Cl^-}$

Total particles $= 1+1+1+2 = 5$ ions? Actually distinct ions formed are: $\mathrm{K^+,\ Mg^{2+},\ 3Cl^-}$ So total number of ions $= 1+1+3 = 5$. Thus $i=5$.

• Hence, $iM = 5 \times 0.02 = 0.10$

2. $0.05\,\text{M } \mathrm{KCl}$
   • $i=2$
   • Hence, $iM = 2 \times 0.05 = 0.10$

So, pair E has same osmotic pressure.

2. Count the matching pairs

Pairs with same osmotic pressure are:

• A
• B
• D
• E

Total number of such pairs:

$4$

3. Comparison with stored answer

Stored correct answer = $4$

Our derived answer = $4$

They match.