Question

For the following question, enter the correct numerical value upto TWO decimal places. If the numerical value has more than two decimal places, round-off the value to TWO decimal places. (For example: Numeric value 5 will be written as 5.00 and 2.346 will be written as 2.35) The van’t Hoff factor for 0.1 M Barium nitrate is 2.74. The percentage of dissociation of Barium nitrate is _____

Hint:

The van’t Hoff factor \(i\) helps determine the degree of dissociation using \(i = 1 + (n-1)\alpha\), where \(n\) is the number of ions formed from one formula unit of the solute.

Answer 1

Correct Answer: 87

Step 1: Recall the relation between van’t Hoff factor (\(i\)) and degree of dissociation (\(\alpha\)) for an electrolyte: \[ i = 1 + (n-1)\alpha \] where \(n\) = number of ions produced per formula unit.
Step 2: Barium nitrate (\(\mathrm{Ba(NO_3)_2}\)) dissociates as: \[ \mathrm{Ba(NO_3)_2 \longrightarrow Ba^{2+} + 2 NO_3^-} \] So, \(n = 3\).
Step 3: Substitute the given \(i = 2.74\) into the formula: \[ i = 1 + (n-1)\alpha \quad \Rightarrow \quad 2.74 = 1 + (3-1)\alpha \]
Step 4: Solve for \(\alpha\): \[ 2.74 - 1 = 2 \alpha \quad \Rightarrow \quad 1.74 = 2 \alpha \quad \Rightarrow \quad \alpha = \frac{1.74}{2} = 0.87 \]
Step 5: Convert \(\alpha\) to percentage: \[ \text{Percentage dissociation} = \alpha \times 100 = 0.87 \times 100 = 87.00% \]