High School

We appreciate your visit to What is the expected boiling point of a solution of 128 g KCl a strong electrolyte dissolved in 1 3 kg of water The molar. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!

What is the expected boiling point of a solution of 128 g KCl (a strong electrolyte) dissolved in 1.3 kg of water?

The molar mass of KCl is 74.55 g/mol, and [tex]K_b = 0.51 \, \text{°C/m}[/tex].

Answer :

Final answer:

The expected boiling point of the solution can be calculated using the formula ΔTb = Kb * m * i, where ΔTb is the boiling point elevation, Kb is the molal boiling point elevation constant, m is the molality of the solution, and i is the van't Hoff factor. For this solution, the expected boiling point is 100.76°C.

Explanation:

The boiling point elevation of a solution can be calculated using the formula ∆Tb = Kb * m * i, where ∆Tb is the boiling point elevation, Kb is the molal boiling point elevation constant, m is the molality of the solution, and i is the van't Hoff factor. In this case, the molality (m) can be calculated by dividing the moles of solute by the mass of the solvent:

m = (moles of solute) / (mass of solvent)

For the given solution, the moles of KCl can be calculated by dividing the mass of KCl by its molar mass:

moles of KCl = (mass of KCl) / (molar mass of KCl)

The final step is to calculate the van't Hoff factor, which depends on the degree of dissociation of the solute in the solvent. KCl is a strong electrolyte and completely dissociates into two ions (K+ and Cl-) in water, so the van't Hoff factor (i) is 2.

Plugging in the values, we can calculate the boiling point elevation:

∆Tb = (Kb) * (m) * (i)

Where:

Kb = 0.51 °C/m

mass of KCl = 128 g

mass of water = 1.3 kg

molar mass of KCl = 74.55 g/mol

i = 2

Substituting the values:

∆Tb = (0.51 °C/m) * [(128 g / 74.55 g/mol) / 1.3 kg] * 2

After calculating, the expected boiling point elevation is ∆Tb = 0.76 °C. To find the expected boiling point of the solution, add the boiling point elevation to the boiling point of pure water, which is 100 °C:

Expected boiling point = 100 °C + 0.76 °C = 100.76 °C

Thanks for taking the time to read What is the expected boiling point of a solution of 128 g KCl a strong electrolyte dissolved in 1 3 kg of water The molar. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!

Rewritten by : Barada