Questions: For each of the following, use appropriate chemical principles to explain the observation. Include chemical equations as appropriate. a. When table salt (NaCl) and sugar (C12H22O11) are dissolved in water, it is observed that i. Both solutions have higher boiling points than pure water ii. The boiling point of 0.10 M NaCl(aq) is higher than that of 0.10 M C12H22O11(aq). b. Water droplets form on the outside of a beaker containing an ice bath.

For each of the following, use appropriate chemical principles to explain the observation. Include chemical equations as appropriate.

a. When table salt (NaCl) and sugar (C12H22O11) are dissolved in water, it is observed that
i. Both solutions have higher boiling points than pure water
ii. The boiling point of 0.10 M NaCl(aq) is higher than that of 0.10 M C12H22O11(aq).

b. Water droplets form on the outside of a beaker containing an ice bath.

Transcript text: 2. For each of the following, use appropriate chemical principles to explain the observation. Include chemical equations as appropriate. a. When table salt $(\mathrm{NaCl})$ and sugar $\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)$ are dissolved in water, it is observed that i. Both solutions have higher boiling points than pure water ii. The boiling point of $0.10 \mathrm{M} \mathrm{NaCl}(\mathrm{aq})$ is higher than that of $0.10 \mathrm{M} \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(\mathrm{aq})$. b. Water droplets form on the outside of a beaker containing an ice bath.

Solution

Solution Steps

Step 1: Understanding Boiling Point Elevation

When a solute is dissolved in a solvent, the boiling point of the resulting solution is higher than that of the pure solvent. This phenomenon is known as boiling point elevation and is a colligative property, meaning it depends on the number of solute particles in the solution, not their identity.

Step 2: Explanation for Part (a)(i)

For both table salt $(\mathrm{NaCl})$ and sugar $\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)$, when dissolved in water, they increase the boiling point of the solution compared to pure water. This is because the presence of solute particles disrupts the formation of vapor bubbles within the liquid, requiring a higher temperature to reach the boiling point.

Step 3: Explanation for Part (a)(ii)

The boiling point elevation is greater for $0.10 \mathrm{M} \mathrm{NaCl}(\mathrm{aq})$ than for $0.10 \mathrm{M} \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(\mathrm{aq})$ because $\mathrm{NaCl}$ dissociates into two ions, $\mathrm{Na}^+$ and $\mathrm{Cl}^-$, in solution. This effectively doubles the number of solute particles compared to sugar, which does not dissociate. Therefore, the boiling point elevation is greater for the $\mathrm{NaCl}$ solution.

Step 4: Explanation for Part (b)

Water droplets form on the outside of a beaker containing an ice bath due to condensation. The ice bath cools the air surrounding the beaker, causing the water vapor in the air to lose energy and condense into liquid water on the surface of the beaker.

Final Answer

a.
i. Both solutions have higher boiling points than pure water due to boiling point elevation, a colligative property.
ii. The boiling point of $0.10 \mathrm{M} \mathrm{NaCl}(\mathrm{aq})$ is higher than that of $0.10 \mathrm{M} \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(\mathrm{aq})$ because $\mathrm{NaCl}$ dissociates into more particles.

b. Water droplets form on the outside of a beaker containing an ice bath due to condensation of water vapor from the air.

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