List the following aqueous solutions in order of increasing boiling point: 0.120 \(\mathrm{m}\) glucose, 0.050 \(\mathrm{m}\) LiBr, 0.050 \(\mathrm{m}\) \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\)

Boiling point elevation is a colligative property, meaning it depends only on the number of solute particles in the solution, not their identity. The formula for boiling point elevation is: \[\Delta T_b = K_b \cdot m \cdot i\] Where \(\Delta T_b\) is the change in boiling point, \(K_b\) is the molal boiling point elevation constant for the solvent (water in this case), \(m\) is the molality of the solute, and \(i\) is the van't Hoff factor (number of particles formed by one formula unit of the solute).

We will calculate the boiling point elevation for each solution using the given molality and respective van't Hoff factors for each solute. For glucose (C₆H₁₂O₆), the formula unit doesn't dissociate in water, so its van't Hoff factor (i) is 1. For LiBr, the formula unit dissociates into two ions (Li⁺ and Br⁻), so its van't Hoff factor (i) is 2. For Zn(NO₃)₂, the formula unit dissociates into three ions (Zn²⁺, 2NO₃⁻), so its van't Hoff factor (i) is 3. Now, we can calculate the boiling point elevation for each solution: \[\Delta T_{b, glucose} = K_b \cdot m_{glucose} \cdot i_{glucose}\] \[\Delta T_{b, LiBr} = K_b \cdot m_{LiBr} \cdot i_{LiBr}\] \[\Delta T_{b, Zn(NO_3)_2} = K_b \cdot m_{Zn(NO_3)_2} \cdot i_{Zn(NO_3)_2}\] Using the given molality and calculated van't Hoff factors: \[\Delta T_{b, glucose} = K_b \cdot 0.120 \cdot 1\] \[\Delta T_{b, LiBr} = K_b \cdot 0.050 \cdot 2\] \[\Delta T_{b, Zn(NO_3)_2} = K_b \cdot 0.050 \cdot 3\]

We can compare the calculated boiling point elevations (note that the \(K_b\) for water will be the same in all cases and doesn't affect the ranking): \[\Delta T_{b, glucose} < \Delta T_{b, LiBr} < \Delta T_{b, Zn(NO_3)_2}\] Therefore, the aqueous solutions are ranked in order of increasing boiling point as follows: 0.120 m glucose < 0.050 m LiBr < 0.050 m Zn(NO₃)₂