Question

Arrange the following equimolar solutions according to increasing order of osmotic pressure [Assume complete ionisation]
i) KCl
ii) $BaCl_2$
iii) $AlCl_3$
iv) $Al_2(SO_4)_3$

• A. $BaCl_2 < Al_2(SO_4)_3 < KCl < AlCl_3$
• B. $Al_2(SO_4)_3 < KCl < BaCl_2 < AlCl_3$
• C. $KCl < BaCl_2 < AlCl_3 < Al_2(SO_4)_3$
• D. $AlCl_3 < BaCl_2 < Al_2(SO_4)_3 < KCl$

Hint:

For 100% ionized strong electrolytes, simply count the subscripts in the chemical formula to find the value of $i$. For example, in $Al_2(SO_4)_3$, 2 aluminums + 3 sulfates = 5 total ions.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Osmotic pressure ($\pi$) is a colligative property given by $\pi = iCRT$. For equimolar solutions ($C, R, T$ are same), $\pi \propto i$ (van't Hoff factor).

Step 2: Formula Application:
Count the number of ions ($i$) produced upon complete dissociation: - (i) KCl $\rightarrow K^+ + Cl^-$ ($i = 2$) - (ii) $BaCl_2 \rightarrow Ba^{2+} + 2Cl^-$ ($i = 3$) - (iii) $AlCl_3 \rightarrow Al^{3+} + 3Cl^-$ ($i = 4$) - (iv) $Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-}$ ($i = 5$)

Step 3: Explanation:
Since $i$ values follow the order $2 < 3 < 4 < 5$, the osmotic pressure follows the order $KCl < BaCl_2 < AlCl_3 < Al_2(SO_4)_3$.

Step 4: Final Answer:
The correct increasing order is Option (C).