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Calculate the value of \(m_{\pm}\) in \(5.5 \times 10^{-3}\) molal solutions of (a) \(\mathrm{KCl}\) (b) \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2},\) and (c) \(\mathrm{ZnSO}_{4}\). Assume complete dissociation.

Short Answer

Expert verified
The value of \(m_{\pm}\) for the given 5.5 脳 10鈦宦 molal solutions is: (a) KCl: \(m_{\pm} = 11 \times 10^{-3}\) molal (b) 颁补(狈翱鈧)鈧: \(m_{\pm} = 16.5 \times 10^{-3}\) molal (c) 窜苍厂翱鈧: \(m_{\pm} = 11 \times 10^{-3}\) molal

Step by step solution

01

Write the dissociation reactions for each compound.

We need to determine the number of ions produced after dissociation for each compound. Let's write down the dissociation reactions for KCl, Ca(NO鈧)鈧, and 窜苍厂翱鈧: (a) KCl 鈫 K鈦 + Cl鈦 (b) Ca(NO鈧)鈧 鈫 Ca虏鈦 + 2NO鈧冣伝 (c) ZnSO鈧 鈫 Zn虏鈦 + SO鈧劼测伝
02

Determine the number of ions produced after dissociation for each compound.

From the dissociation reactions in Step 1, we can see the number of ions produced for each compound: (a) KCl produces 1 K鈦 and 1 Cl鈦 ions, giving a total of 2 ions. (b) Ca(NO鈧)鈧 produces 1 Ca虏鈦 and 2 NO鈧冣伝 ions, giving a total of 3 ions. (c) ZnSO鈧 produces 1 Zn虏鈦 and 1 SO鈧劼测伝 ions, giving a total of 2 ions.
03

Calculate the value of molality (m鈧娾倠) for KCl, Ca(NO鈧)鈧, and ZnSO鈧.

We can now calculate the value of m鈧娾倠 by multiplying the number of ions produced (from Step 2) with the given molality of the solution (5.5 脳 10鈦宦 molal). For KCl: m鈧娾倠 = (Number of ions) 脳 (Molality of solution) m鈧娾倠 = 2 脳 5.5 脳 10鈦宦 m鈧娾倠 = 11 脳 10鈦宦 molal For 颁补(狈翱鈧)鈧: m鈧娾倠 = (Number of ions) 脳 (Molality of solution) m鈧娾倠 = 3 脳 5.5 脳 10鈦宦 m鈧娾倠 = 16.5 脳 10鈦宦 molal For 窜苍厂翱鈧: m鈧娾倠 = (Number of ions) 脳 (Molality of solution) m鈧娾倠 = 2 脳 5.5 脳 10鈦宦 m鈧娾倠 = 11 脳 10鈦宦 molal After calculating the value of m鈧娾倠 for each solution, we get: (a) KCl: m鈧娾倠 = 11 脳 10鈦宦 molal (b) 颁补(狈翱鈧)鈧: m鈧娾倠 = 16.5 脳 10鈦宦 molal (c) 窜苍厂翱鈧: m鈧娾倠 = 11 脳 10鈦宦 molal

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Key Concepts

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

Molality
Molality is a measure of the concentration of a solute in a solution. It is expressed as the number of moles of solute per kilogram of solvent. Unlike molarity, which is affected by temperature and pressure due to volume changes, molality remains consistent because it is based on mass.
To calculate molality (\[ m \]), you should follow these steps:
  • Determine the number of moles of the solute present.
  • Measure the mass of the solvent in kilograms.
The formula for molality is expressed as:
\[ m = \frac{{ ext{{moles of solute}}}}{{ ext{{kilograms of solvent}}}} \]
Understanding molality is crucial for calculations involving colligative properties, which depend on the concentration of solute particles rather than their identity.
Ion Counting
Ion counting is the process of determining the number of ions produced when a compound dissociates in a solution. This is an essential step in understanding ionic compounds' behavior in solutions.
To illustrate, consider these examples:
  • KCl: On dissociation, it forms one \( ext{K}^+ \) ion and one \( ext{Cl}^- \) ion, totaling two ions.
  • 颁补(狈翱鈧)鈧: It dissociates into one \( ext{Ca}^{2+} \) ion and two \( ext{NO}_3^- \) ions, giving three ions in total.
  • 窜苍厂翱鈧: Results in one \( ext{Zn}^{2+} \) ion and one \( ext{SO}_4^{2-} \) ion, again giving two ions in total.
Understanding ion counting is key in calculating properties like ionic strength or effective concentration, useful in various fields such as chemistry and environmental science.
Molal Solutions
Molal solutions refer to those prepared by dissolving a known amount of solute in a specific mass of solvent, expressed in molality. An important distinction from molar solutions, which rely on a fixed volume of the solution.
Creating a molal solution involves:
  • Selecting the solute and determining the number of moles required.
  • Weighing the solvent precisely.
  • Ensuring complete dissolution of the solute in the solvent.
When dealing with compounds that dissociate into ions, such as salts, calculating the effective concentration is crucial. For example, in a 5.5 x \( 10^{-3} \) molal solution of \( ext{KCl} \) or \( ext{ZnSO}_4 \), accounting for dissociation results in an effective concentration of roughly 11 x \( 10^{-3} \) molal. However, compounds like \( ext{Ca(NO}_3 ext{)}_2 \) generate more ions, increasing the effective concentration to \( 16.5 imes 10^{-3} \) molal. Understanding molal solutions is critical in various chemical calculations, especially those involving colligative properties.

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Most popular questions from this chapter

Estimate the degree of dissociation of a \(0.200 \mathrm{m}\) solution of nitrous acid \(\left(K_{a}=4.00 \times 10^{-4}\right)\) that is also \(0.500 \mathrm{m}\) in the strong electrolyte given in parts (a) through (c). Use the data tables to obtain \(\gamma_{\pm},\) as the electrolyte concentration is too high to use the Debye-H眉ckel limiting law. a. \(\mathrm{Ba}(\mathrm{Cl})_{2}\) b. \(\mathrm{KOH}\) \(\mathbf{c} . \mathrm{AgNO}_{3}\) Compare your results with the degree of dissociation of the acid in the absence of other electrolytes.

At \(25^{\circ} \mathrm{C},\) the equilibrium constant for the dissociation of acetic acid \(K_{a}\) is \(1.75 \times 10^{-5}\). Using the Debye-H眉ckel limiting law, calculate the degree of dissociation in \(0.150 \mathrm{m}\) and \(1.50 \mathrm{m}\) solutions using an iterative calculation until the answer is constant to within \(+/-2\) in the second decimal place. Compare these values with what you would obtain if the ionic interactions had been ignored. Compare your results with the degree of dissociation of the acid assuming \(\gamma_{\pm}=1\)

Calculate the mean ionic molality and mean ionic activity of a \(0.105 \mathrm{m} \mathrm{K}_{3} \mathrm{PO}_{4}\) solution for which the mean ionic activity coefficient is 0.225

Calculate the pH of a buffer solution that is 0.200 molal in \(\mathrm{CH}_{3} \mathrm{COOH}\) and 0.15 molal in \(\mathrm{CH}_{3} \mathrm{COONa}\) using the Davies equation to calculate \(\gamma_{\pm} .\) What pH value would you have calculated if you had assumed that \(\gamma_{\pm}=1 ?\)

Use the Davies equation to calculate \(\gamma_{\pm}\) for a 1.00 molar solution of KOH. Compare your answer with the value in Table 10.3

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