The Van't Hoff factor for 0.1M $Ba(NO_{3})_{2}$ solution is 2.74, then degree of dissociation is :

Q: The Van't Hoff factor for 0.1M $Ba(NO_{3})_{2}$ solution is 2.74, then degree of dissociation is :

To determine the degree of dissociation ($\alpha$) of $Ba(NO_3)_2$ in a 0.1M solution, we use the Van't Hoff factor ($i$) and the dissociation equation. The Van't Hoff factor for the solution is given as 2.74. ### Explanation: 1. **Understanding the Dissociation:** - $Ba(NO_3)_2$ dissociates in water as follows: \[ Ba(NO_3)_2 \rightarrow Ba^{2+} + 2NO_3^- \] - Ideally, if the dissociation were complete, one formula unit of $Ba(NO_3)_2$ would produce 3 ions (1 $Ba^{2+}$ and 2 $NO_3^-$), so the theoretical Van't Hoff factor ($i$) would be 3. 2. **Using the Van't Hoff Factor:** - The Van't Hoff factor ($i$) is related to the degree of dissociation ($\alpha$) by the formula: \[ i = 1 + \alpha(n - 1) \] where $n$ is the number of ions produced from one formula unit of the solute. For $Ba(NO_3)_2$, $n = 3$. 3. **Calculating the Degree of Dissociation ($\alpha$):** - Plugging the values into the formula: \[ 2.74 = 1 + \alpha(3 - 1) \] \[ 2.74 = 1 + 2\alpha \] \[ 2.74 - 1 = 2\alpha \] \[ 1.74 = 2\alpha \] \[ \alpha = \frac{1.74}{2} = 0.87 \] 4. **Conclusion:** - The calculated degree of dissociation ($\alpha$) is 0.87, which corresponds to Option 3. ### Explanation of Incorrect Options: - **Option 1 (1):** This would imply complete dissociation, which would result in a Van't Hoff factor of 3, not 2.74. Therefore, this option is incorrect. - **Option 2 (0.913):** This value would result in a Van't Hoff factor greater than 2.74 when plugged into the formula, making it inconsistent with the given data. - **Option 4 (0.74):** This value would result in a Van't Hoff factor less than 2.74, which does not match the given information. Thus, the correct answer is Option 3, with a degree of dissociation of 0.87.

Question

Accepted Answer

To determine the degree of dissociation ($\alpha$) of $Ba(NO_3)_2$ in a 0.1M solution, we use the Van't Hoff factor ($i$) and the dissociation equation. The Van't Hoff factor for the solution is given as 2.74.

### Explanation:

1. **Understanding the Dissociation:**
   - $Ba(NO_3)_2$ dissociates in water as follows:
     $$
     Ba(NO_3)_2 \rightarrow Ba^{2+} + 2NO_3^-
     $$
   - Ideally, if the dissociation were complete, one formula unit of $Ba(NO_3)_2$ would produce 3 ions (1 $Ba^{2+}$ and 2 $NO_3^-$), so the theoretical Van't Hoff factor ($i$) would be 3.

2. **Using the Van't Hoff Factor:**
   - The Van't Hoff factor ($i$) is related to the degree of dissociation ($\alpha$) by the formula:
     $$
     i = 1 + \alpha(n - 1)
     $$
     where $n$ is the number of ions produced from one formula unit of the solute. For $Ba(NO_3)_2$, $n = 3$.

3. **Calculating the Degree of Dissociation ($\alpha$):**
   - Plugging the values into the formula:
     $$
     2.74 = 1 + \alpha(3 - 1)
     $$
     $$
     2.74 = 1 + 2\alpha
     $$
     $$
     2.74 - 1 = 2\alpha
     $$
     $$
     1.74 = 2\alpha
     $$
     $$
     \alpha = \frac{1.74}{2} = 0.87
     $$

4. **Conclusion:**
   - The calculated degree of dissociation ($\alpha$) is 0.87, which corresponds to Option 3.

### Explanation of Incorrect Options:

- **Option 1 (1):** This would imply complete dissociation, which would result in a Van't Hoff factor of 3, not 2.74. Therefore, this option is incorrect.
  
- **Option 2 (0.913):** This value would result in a Van't Hoff factor greater than 2.74 when plugged into the formula, making it inconsistent with the given data.

- **Option 4 (0.74):** This value would result in a Van't Hoff factor less than 2.74, which does not match the given information.

Thus, the correct answer is Option 3, with a degree of dissociation of 0.87.