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(a) Does a 0.10 \(\mathrm{m}\) aqueous solution of NaCl have a higher bolling point, a lower boiling point, or the same boiling point as a 0.10 \(\mathrm{m}\) aqueous solution of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} ?(\mathbf{b})\) The experimental boiling point of the NaCl solution is lower than that calculated assuming that NaCl is completely dissociated in solution. Why is this the case?

Short Answer

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(a) A 0.10 M aqueous solution of NaCl will have a higher boiling point than a 0.10 M aqueous solution of C6H12O6, as NaCl has a greater van't Hoff factor (2) than C6H12O6 (1). (b) The lower experimental boiling point for the NaCl solution is due to some Na+ and Cl- ions being associated (ion pairing) in the solution rather than being completely dissociated. This reduces the van't Hoff factor (i) to a value less than 2, leading to a smaller boiling point elevation.

Step by step solution

01

Comparing Boiling Points of NaCl and C6H12O6 Solutions

First, let's recall the boiling point elevation equation: \[ΔT_b = i × K_b × m\] Where ΔT_b is the boiling point elevation, i is the van't Hoff factor, K_b is the boiling point elevation constant, and m is the molality. For NaCl, i = 2 (due to its complete dissociation into Na+ and Cl- ions). For C6H12O6, i = 1 (no dissociation, as it is a non-electrolyte). Given that both solutions have the same molality (0.10 M), we can compare their boiling point elevations directly by comparing their van't Hoff factors. Since the van't Hoff factor for NaCl is greater than that for C6H12O6 (2 > 1), the boiling point elevation for the NaCl solution will be higher. Thus, a 0.10 M aqueous solution of NaCl will have a higher boiling point than a 0.10 M aqueous solution of C6H12O6.
02

Explaining the Lower Experimental Boiling Point for NaCl Solution

The assumption that NaCl is completely dissociated in the solution leads to a higher calculated boiling point because the van't Hoff factor (i) will be 2. However, in reality, some Na+ and Cl- ions may be associated (ion pairing) in solution rather than being completely dissociated. This would effectively reduce the van't Hoff factor (i) to a value less than 2. Since the van't Hoff factor (i) in the boiling point elevation equation is directly proportional to the boiling point elevation, a smaller value for i will lead to a smaller value for the boiling point elevation. As a result, the experimental boiling point is lower than what would be calculated assuming complete dissociation of NaCl.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

van't Hoff factor
The van't Hoff factor, represented by the symbol 'i', is significant in understanding boiling point elevation among other colligative properties. It quantifies the number of particles into which a compound dissociates in solution. For instance, NaCl splits into two particles, Na+ and Cl-, giving it a van't Hoff factor of 2. In contrast, glucose (\text{C}_6 \text{H}_{12} \text{O}_6), being a non-electrolyte, does not dissociate, so its factor remains 1.In boiling point elevation, the higher the van't Hoff factor, the more significant the impact on the solution’s boiling point. The factor is included in the formula \[ΔT_b = i × K_b × m\] and helps determine the actual effect of solutes on the solution's boiling point. Real-world scenarios, however, may reflect a lower van't Hoff factor for salts like NaCl due to ion association, which decreases their expected impact on boiling point elevation and explains why experimental data might differ from theoretical calculations.
molality
Molality, denoted as 'm', measures the concentration of a solute in a solvent. Unlike molarity, molality is based on the mass of the solvent, not the volume of the solution, making it temperature-independent. It is defined as the number of moles of solute per kilogram of solvent. In our exercise, the molality of both NaCl and glucose solutions is 0.10 mol/kg.When we delve into boiling point elevation, molality becomes a crucial factor, as it is directly proportional to the increase in boiling point according to the formula \(ΔT_b = i × K_b × m\). Therefore, at the same molality, different substances can affect the boiling point differently due to their respective van’t Hoff factors.
colligative properties
Colligative properties are characteristics of a solution that depend on the ratio of the number of solute particles to the number of solvent molecules in a solution, and not on the nature of the chemical species present. Boiling point elevation is one such property, along with freezing point depression, vapor pressure lowering, and osmotic pressure.Colligative properties are essential in understanding everyday phenomena, such as why salt is spread on icy roads (to lower the freezing point) or how a car's radiator fluid prevents the engine from overheating (boiling point elevation). Specifically, boiling point elevation is the process where the boiling point of a solvent increases when a solute is dissolved in it. The ability to manipulate the boiling points of solutions through these properties is fundamental in various industries, such as food processing and pharmaceuticals, and also in daily cooking practices.

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Most popular questions from this chapter

Adrenaline is the hormone that triggers the release of extra glucose molecules in times of stress or emergency. A solution of 0.64 g of adrenaline in 36.0 g of \(\mathrm{CCl}_{4}\) elevates the boiling point by \(0.49^{\circ} \mathrm{C}\) Calculate the approximate molar mass of adrenaline from this data.

At ordinary body temperature \(\left(37^{\circ} \mathrm{C}\right),\) the solubility of \(\mathrm{N}_{2}\) in water at ordinary atmospheric pressure \((1.0 \mathrm{atm})\) is 0.015 \(\mathrm{g} / \mathrm{L} .\) Air is approximately 78 \(\mathrm{mol} \% \mathrm{N}_{2}\) . (a) Calculate the number of moles of \(\mathrm{N}_{2}\) dissolved per liter of blood, assuming blood is a simple aqueous solution. (b) At a depth of 100 \(\mathrm{ft}\) in water, the external pressure is 4.0 atm. What is the solubility of \(\mathrm{N}_{2}\) from air in blood at this pressure? (c) If a scuba diver suddenly surfaces from this depth, how many milliliters of \(\mathrm{N}_{2}\) gas, in the form of tiny bubbles, are released into the bloodstream from each liter of blood?

Indicate whether each statement is true or false: (a) The higher the temperature, the more soluble most gases are in water. (b) The higher the temperature, the more soluble most ionic solids are in water. (c) As you cool a saturated solution from high temperature to low temperature, solids start to crystallize out of solution if you achieve a supersaturated solution. (d) If you take a saturated solution and raise its temperature, you can (usually) add more solute and make the solution even more concentrated.

The concentration of gold in seawater has been reported to be between 5 ppt (parts per trillion) and 50 ppt. Assuming that seawater contains 13 ppt of gold, calculate the number of grams of gold contained in \(1.0 \times 10^{3}\) gal of seawater.

Glucose makes up about 0.10\(\%\) by mass of human blood. Calculate this concentration in (a) ppm, (b) molality. (c) What further information would you need to determine the molarity of the solution?

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