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 Question:
We are given four electrolyte solutions at identical (equimolar) concentrations and instructed to assume they undergo 100% dissociation.
We must rank them in increasing order of their theoretical osmotic pressures.

Step 2: Key Formula or Approach:
Osmotic pressure ($\pi$) is a strictly colligative property, meaning it depends exclusively on the total number of solute particles present in the solution, not their chemical identity.
The formula is:
$$\pi = iCRT$$
Since concentration ($C$), the gas constant ($R$), and temperature ($T$) are constant across all equimolar solutions, the osmotic pressure becomes directly proportional to the van't Hoff factor ($i$):
$$\pi \propto i$$

Step 3: Detailed Explanation:
We must find the van't Hoff factor ($i$) for each fully dissociating salt by counting the number of ions produced per formula unit:
i) $KCl \rightarrow K^+ + Cl^-$ (yields 2 ions, so $i = 2$)
ii) $BaCl_2 \rightarrow Ba^{2+} + 2Cl^-$ (yields 3 ions, so $i = 3$)
iii) $AlCl_3 \rightarrow Al^{3+} + 3Cl^-$ (yields 4 ions, so $i = 4$)
iv) $Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-}$ (yields 5 ions, so $i = 5$)
Since $\pi \propto i$, the order of osmotic pressure directly mirrors the order of the van't Hoff factors:
$2 < 3 < 4 < 5$
Therefore: $KCl < BaCl_2 < AlCl_3 < Al_2(SO_4)_3$.

Step 4: Final Answer:
The correct increasing order is shown in option (c).