Which solutions will have the highest boiling point?

1 M \(BaCl_2\) solution

The correct answer is option 1. 1 M \(BaCl_2\) solution.

The Core Principle

The boiling point of a solution ($T_b$) increases with the number of solute particles present in the solution. This is known as elevation in boiling point ($\Delta T_b$).

• The formula is: $\Delta T_b = i \cdot K_b \cdot m$
• Since the molarity (1 M) is the same for all options, the boiling point depends entirely on the van't Hoff factor ($i$), which is the number of ions/particles produced after dissociation.

We calculate the number of particles ($i$) for each option assuming complete dissociation:

1. $BaCl_2$ (Barium Chloride): Dissociates into $1$ $Ba^{2+}$ ion and $2$ $Cl^-$ ions.
• $i = 3$ (Total $3$ particles)
2. $NaCl$ (Sodium Chloride): Dissociates into $1$ $Na^+$ ion and $1$ $Cl^-$ ion.
• $i = 2$ (Total $2$ particles)
3. $C_6H_{12}O_6$ (Glucose): It is a non-electrolyte and does not dissociate.
• $i = 1$ (Total $1$ particle)
4. $(NH_2)_2CO$ (Urea): It is also a non-electrolyte and does not dissociate.
• $i = 1$ (Total $1$ particle)

Highest particles ($i = 3$): $BaCl_2$

Result: It causes the maximum elevation in boiling point ($\Delta T_b$).

Boiling Point Order: $BaCl_2 > NaCl > \text{Glucose} = \text{Urea}$