We have four aqueous solutions labelled as:
A. \(0.1\, \ M\, \ NaCl\)
B. \(0.01\, \ M\, \ NaCl\)
C. \(0.01\, \ M\, \ BaCl_2\)
D. \(0.01\, \ M\, \ Sucrose\)
Choose the correct increasing order of van't Hoff factor from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is option 3. \(i_D < i_A < i_B < i_C\).
In real solutions, the observed van't Hoff factor is slightly less than the theoretical value because of interionic attractions. As a solution becomes more concentrated, these attractions increase, causing the degree of dissociation to decrease.
- $i_A$ vs $i_B$: Since Solution A ($0.1\ M$) is more concentrated than Solution B ($0.01\ M$), its observed van't Hoff factor ($i_A$) will be slightly lower than $i_B$.
Increasing Order Comparison
- Lowest ($i_D$): Sucrose is a non-electrolyte ($i = 1$).
- Middle-Low ($i_A$): $NaCl$ at higher concentration ($0.1\ M$) has more interionic attraction.
- Middle-High ($i_B$): $NaCl$ at lower concentration ($0.01\ M$) is more dissociated.
- Highest ($i_C$): $BaCl_2$ produces 3 ions per formula unit.
Therefore, the correct increasing order is:
$i_D < i_A < i_B < i_C$