Which of these solutes raises the boiling point of water the most?

Question

A.
ionic lithium chloride (LiCl)
B.
ionic sodium chloride (NaCl)
C.
molecular sucrose (C12H22O11)
D.
ionic sodium phosphate (Na3PO4)
E.
ionic magnesium bromide (MgBr2)

Answer 1

To determine which solute raises the boiling point of water the most, we can use the principle of colligative properties, which states that the boiling point elevation of a solvent is related to the number of solute particles in solution. The formula for boiling point elevation is:

$$\mathrm{\Delta }{T}_b=i\cdot K_b\cdot m$$

Where:

• $\mathrm{\Delta }{T}_b$ is the boiling point elevation,
• $i$ is the van 't Hoff factor (the number of particles the compound dissociates into),
• $K_b$ is the ebullioscopic constant of the solvent (water in this case),
• $m$ is the molality of the solution.

Now, let's look at each option:

A. Lithium Chloride (LiCl): Dissociates into 2 particles (Li⁺ and Cl⁻), so $i=2$.

B. Sodium Chloride (NaCl): Dissociates into 2 particles (Na⁺ and Cl⁻), so $i=2$.

C. Sucrose (C12H22O11): Does not dissociate, so $i=1$.

D. Sodium Phosphate (Na3PO4): Dissociates into 4 particles (3 Na⁺ and 1 PO4³⁻), so $i=4$.

E. Magnesium Bromide (MgBr2): Dissociates into 3 particles (1 Mg²⁺ and 2 Br⁻), so $i=3$.

Now, comparing the values of $i$:

• LiCl: $i=2$
• NaCl: $i=2$
• Sucrose: $i=1$
• Na3PO4: $i=4$
• MgBr2: $i=3$

Since sodium phosphate (Na3PO4) has the highest van 't Hoff factor ($i=4$), it will raise the boiling point of water the most.

The answer is D. ionic sodium phosphate (Na3PO4).