Q39 / 180 · COMEDK 2024 (Morning Shift) · Chemistry
$\text { If a compound } \mathrm{X}_3 \mathrm{Y} \text { is } 60 \% \text { ionised in aqueous medium, what is its van't Hoff factor value? }$
- A3.1
- B0.8
- C0.6
- Correct answerD2.8
Solution
The compound $X_3Y$ dissociates in aqueous medium as follows:
$X_3Y \rightarrow 3X^{+} + Y^{3-}$
The number of ions produced per formula unit is $n = 3 + 1 = 4$.
The degree of dissociation is given as $\alpha = 60\% = 0.6$.
The van't Hoff factor $i$ is calculated using the formula $i = 1 + \alpha(n - 1)$.
Substituting the values $n = 4$ and $\alpha = 0.6$ into the formula:
$i = 1 + 0.6(4 - 1)$
$i = 1 + 0.6(3)$
$i = 1 + 1.8$
$i = 2.8$
Answer: 2.8
$X_3Y \rightarrow 3X^{+} + Y^{3-}$
The number of ions produced per formula unit is $n = 3 + 1 = 4$.
The degree of dissociation is given as $\alpha = 60\% = 0.6$.
The van't Hoff factor $i$ is calculated using the formula $i = 1 + \alpha(n - 1)$.
Substituting the values $n = 4$ and $\alpha = 0.6$ into the formula:
$i = 1 + 0.6(4 - 1)$
$i = 1 + 0.6(3)$
$i = 1 + 1.8$
$i = 2.8$
Answer: 2.8
Hence, the correct answer is option: 'D'.