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What is the boiling point of a 0.10\(M\) solution of \(\mathrm{NaHSO}_{4}\) if the solution has a density of 1.002 \(\mathrm{g} / \mathrm{mL} ?\)

Short Answer

Expert verified
The boiling point of a 0.10 M solution of NaHSO鈧 with a density of 1.002 g/mL is approximately 100.00854掳C. To find this value, we used the molality, van't Hoff factor, and molal boiling point elevation constant in the boiling point elevation formula.

Step by step solution

01

Find the molal concentration (molality) of the solution.

To find the molal concentration of the solution, we need to use the molar concentration (M) and the density of the solution: Molality = (Molarity 脳 Density) / (1 + Molarity 脳 Molar_Mass) Given: Molarity (M) = 0.10 M Density = 1.002 g/mL Molar mass of NaHSO鈧 can be calculated as: Na = 22.99 g/mol H = 1.01 g/mol S = 32.07 g/mol O鈧 = 64 g/mol Molar_Mass = 22.99 + 1.01 + 32.07 + 64 = 120.07 g/mol Now, substitute values and find the molality of the solution: Molality = (0.10 脳 1.002) / (1 + 0.10 脳 120.07) Molality = 0.1002 / (1 + 12.007) Molality 鈮 0.00834 mol/kg
02

Determine the van't Hoff factor (i).

The van't Hoff factor (i) represents how many particles the solute dissociates into when it dissolves. For NaHSO鈧, which is a strong electrolyte, it will dissociate into two ions: Na鈦 and HSO鈧勨伝. Thus, the van't Hoff factor (i) = 2.
03

Determine the molal boiling point elevation constant (Kb) for water.

The molal boiling point elevation constant (Kb) is a property of the solvent (water in this case) that affects the boiling point elevation. For water, Kb = 0.512 掳C/molal.
04

Calculate the boiling point elevation (螖T).

Now, we can calculate the boiling point elevation (螖T) using the molality, the van't Hoff factor, and the molal boiling point elevation constant: 螖T = i 脳 molality 脳 Kb Substitute the values: 螖T = 2 脳 0.00834 脳 0.512 螖T 鈮 0.00854掳C
05

Determine the actual boiling point of the solution.

Finally, to find the boiling point of the solution, we need to add the boiling point elevation (螖T) to the boiling point of pure water (100掳C): Boiling_Point = 100 + 螖T Boiling_Point 鈮 100 + 0.00854 Boiling_Point 鈮 100.00854掳C Thus, the boiling point of the 0.10 M NaHSO鈧 solution is approximately 100.00854掳C.

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Key Concepts

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

Molality
Understanding the concept of molality is crucial when studying boiling point elevation. Molality is a measure of the concentration of a solute in a solution, defined as the number of moles of solute per kilogram of solvent. Unlike molarity, which is affected by changes in volume due to temperature or pressure, molality remains constant.

To calculate molality, you can use the equation \[ \text{molality} = \frac{\text{moles of solute}}{\text{kilograms of solvent}} \]

This value is particularly useful because it's independent of temperature, making it a reliable unit for studies involving temperature changes, such as in the exercise where you're looking to find the boiling point of a solution.
Van't Hoff Factor
Moving on to the van't Hoff factor, denoted as 'i', it is an important aspect of the boiling point elevation. It accounts for the number of particles into which a solute dissociates in solution. This factor is essential for calculating the boiling point elevation because it helps in determining the actual number of particles that will affect the colligative properties of a solution.

In the given exercise, \[ \text{van't Hoff factor (i)} = 2 \]

because sodium bisulfate (\(\mathrm{NaHSO}_{4}\)) dissociates into two ions 鈥 sodium (\(\mathrm{Na}^{+}\)) and bisulfate (\(\mathrm{HSO}_{4}^{-}\)). Thus, the presence of more particles causes a greater impact on the boiling point of the solution.
Colligative Properties
The term 'colligative properties' refers to the physical properties of solutions that depend on the number of dissolved particles and not on their identity. Boiling point elevation is one such colligative property. Others include freezing point depression, vapor pressure lowering, and osmotic pressure.

When a nonvolatile solute is dissolved in a solvent, such as water, the boiling point of the resulting solution will be higher than that of the pure solvent. This is because the added particles disrupt the solvent's ability to evaporate, requiring more energy (and thus a higher temperature) to reach boiling.
Electrolyte Dissociation
Electrolyte dissociation plays a substantial role in the behavior of solutions. When an electrolyte like sodium bisulfate is dissolved in water, it separates into ions. This dissociation process is crucial to understand because it alters the number of particles in the solution and thereby its colligative properties.

The degree of dissociation is considered when calculating the van't Hoff factor, which is subsequently used to determine the boiling point elevation. Electrolytes can be strong or weak, with strong electrolytes like \(\mathrm{NaHSO}_{4}\) almost completely dissociating into their constituent ions, significantly influencing the solution's boiling point.

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Most popular questions from this chapter

(a) Which of the following is the stronger Bronsted-Lowry acid, \(\mathrm{HClO}_{3}\) or \(\mathrm{HClO}_{2} ?\) (b) Which is the stronger Bronsted-Lowry base, \(\mathrm{HS}^{-}\) or \(\mathrm{HSO}_{4}^{-}\) ?

Consider two solutions, solution \(\mathrm{A}\) and solution \(\mathrm{B} .\left[\mathrm{H}^{+}\right]\) in solution \(\mathrm{A}\) is 250 times greater than that in solution B. What is the difference in the pH values of the two solutions?

Pyridinium bromide \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHBr}\right)\) is a strong electrolyte that dissociates completely into \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\) and \(\mathrm{Br}^{-} .\) An aqueous solution of pyridinium bromide has a pH of \(2.95 .\) (a) Write out the reaction that leads to this acidic pH. (b) Using Appendix D, calculate the \(K_{a}\) for pyridinium bromide. (c) A solution of pyridinium bromide has a pH of 2.95 . What is the concentration of the pyridinium cation atequilibrium, in units of molarity?

Calculate \(\left[\mathrm{OH}^{-}\right]\) and \(\mathrm{pH}\) for (a) \(1.5 \times 10^{-3} \mathrm{MSr}(\mathrm{OH})_{2}\) (b) 2.250 \(\mathrm{g}\) of LiOH in 250.0 \(\mathrm{mL}\) of solution, \((\mathbf{c}) 1.00\) mL of 0.175 M NaOH diluted to \(2.00 \mathrm{L},\) (d) a solution formed by adding 5.00 \(\mathrm{mL}\) of 0.105 \(\mathrm{M} \mathrm{KOH}\) to 15.0 \(\mathrm{mL}\) of \(9.5 \times 10^{-2} \mathrm{MCa}(\mathrm{OH})_{2}.\)

At \(50^{\circ} \mathrm{C},\) the ion-product constant for \(\mathrm{H}_{2} \mathrm{O}\) has the value \(K_{w}=5.48 \times 10^{-14}\) . (a) What is the pH of pure water at \(50^{\circ} \mathrm{C} ?\) (b) Based on the change in \(K_{w}\) with temperature, predict whether \(\Delta H\) is positive, negative, or zero for the autoionization reaction of water: $$2 \mathrm{H}_{2} \mathrm{O}(l) \Longrightarrow \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q)$$

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