The Van't Hoff factor for 0.1 M Ba(NO₃)₂ solution is 2.74. The degree of dissociation is:

Question

Accepted Answer

The relationship between the Van't Hoff factor ($i$), the degree of dissociation ($\alpha$), and the number of ions produced per formula unit ($n$) is given by the formula:
$$i = 1 + (n - 1)\alpha$$

1.  **Identify $n$:**
    *   Barium nitrate dissociates as: $Ba(NO_3)_2 ightarrow Ba^{2+} + 2NO_3^-$.
    *   Total ions ($n$) = $1 + 2 = 3$.

2.  **Substitute values:**
    *   Given $i = 2.74$.
    *   $2.74 = 1 + (3 - 1)\alpha$
    *   $2.74 = 1 + 2\alpha$

3.  **Solve for $\alpha$:**
    *   $2\alpha = 2.74 - 1$
    *   $2\alpha = 1.74$
    *   $\alpha = \frac{1.74}{2} = 0.87$

4.  **Convert to percentage:**
    *   Percentage degree of dissociation = $0.87 	imes 100\% = 87\%$.

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