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Chapter 16: Problem 8

What would be the molality of \(20 \%\) (mass/mass) aqueous solution of \(\mathrm{Kl} ?\) (molar mass of \(\left.\mathrm{Kl}=166 \mathrm{~g} \mathrm{~mol}^{-1}\right)\) (a) \(1.08\) (b) \(1.35\) (c) \(1.48\) (d) \(1.51\)

Short Answer

Expert verified
The molality is approximately 1.51 mol/kg; hence, the answer is (d).

Step by step solution

01

Understand the Problem

We are given a 20% (mass/mass) solution of KI, which means there are 20 grams of KI in every 100 grams of aqueous solution. We need to find the molality of this solution.
02

Identify Given Values

- Mass of KI = 20 g - Mass of solution = 100 g - Molar mass of KI = 166 g/mol
03

Calculate the Mass of Solvent

Since the solution is 100 g in total and contains 20 g of KI, the mass of the solvent (water) is 80 g.
04

Convert Solvent Mass to Kilograms

Convert the mass of the solvent from grams to kilograms: \[80 \, \text{g} \times \frac{1 \, \text{kg}}{1000 \, \text{g}} = 0.08 \, \text{kg}\]
05

Calculate Moles of KI

Calculate the moles of KI using its molar mass:\[\text{Moles of KI} = \frac{20 \, \text{g}}{166 \, \text{g/mol}} \approx 0.1205 \, \text{mol}\]
06

Compute Molality

Molality (m) is calculated as moles of solute per kilogram of solvent:\[m = \frac{0.1205 \, \text{mol}}{0.08 \, \text{kg}} \approx 1.51 \, \text{mol/kg}\]
07

Select the Correct Answer

The calculated molality is approximately 1.51, which corresponds to option (d).

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Key Concepts

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

Molar Mass
Molar mass is a fundamental concept in chemistry that helps us understand the mass of a substance at the molecular level. It tells us how much one mole of a substance weighs, which is crucial for converting between mass and moles.

In our problem, the molar mass of potassium iodide (KI) is given as 166 grams per mole. This means that one mole of KI weighs 166 grams. Having this information allows us to determine how many moles of KI are present in a given mass.

To calculate the moles from the mass, you use the formula:
  • Moles of substance = \( \frac{\text{mass of substance}}{\text{molar mass}} \)
This conversion is critical when working with solutions because it allows us to relate the amount of solute present to its mass, which is often how measurements are made in laboratories.
Mass/Mass Solution
Understanding mass/mass solutions is essential for dealing with concentration in chemistry. A mass/mass solution, denoted as % (m/m), expresses the concentration of a solute in a solution as a percentage by mass.

In our example, a 20% mass/mass solution of KI means 20 grams of KI are dissolved in every 100 grams of the total solution. To get this percentage, the mass of the solute is divided by the total mass of the solution, and then multiplied by 100.

This percentage helps chemists quickly communicate how concentrated a solution is. In practical terms, it also helps in preparing solutions by determining how much solute needs to be added to reach a desired concentration.

Calculation of % mass/mass:
  • % mass/mass = \( \frac{\text{mass of solute}}{\text{total mass of solution}} \times 100 \)
Understanding this concept is key to solving problems related to solution concentration, like finding the mass of the solvent or the solute needed.
Moles of Solute
Moles of solute is a concept that ties the other elements of our calculation together. In any solution, the solute is the substance that is dissolved. The moles of solute tell us how much of the solute is present in the solution in terms of number of particles.

To find the moles of solute, we use the substance's mass and molar mass. In the context of the exercise, you first find the mass of KI, which is 20 grams, and use its molar mass to find the moles.

The relationship between mass of solute and moles is given by the equation:
  • Moles of solute = \( \frac{\text{mass of solute}}{\text{molar mass of solute}} \)
This step is essential because it allows you to compute properties of the solution, such as molality, which relies on knowing how many moles of solute are dissolved in a specific amount of solvent. Hence, it is the crux of quantitatively analyzing solutions in chemistry.

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Most popular questions from this chapter

\(1.22 \mathrm{~g}\) of benzoic acid is dissolved in \(100 \mathrm{~g}\) of acetone and \(100 \mathrm{~g}\) of benzene separately. Boiling point of the solution in acetone increases by \(0.17^{\circ} \mathrm{C}\), while that in the benzene increases by \(0.13^{\circ} \mathrm{C}\); \(K_{b}\) for acetone and benzene is \(1.7 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\) and \(2.6 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\) respectively. Find molecular weight of benzoic acid in two cases and justify your answer.

A solution is prepared by dissolving \(0.6 \mathrm{~g}\) of urea (molar mass \(=60 \mathrm{~g}\) \(\mathrm{mol}^{-1}\) ) and \(1.8 \mathrm{~g}\) of glucose (molar mass \(=180 \mathrm{~g} \mathrm{~mol}^{-1}\) ) in \(100 \mathrm{~mL}\), of water at \(27^{\circ} \mathrm{C}\). The osmotic pressure of the solution is : \(\left(\mathrm{R}=0.08206 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)\) (a) \(8.2 \mathrm{~atm}\) (b) \(2.46 \mathrm{~atm}\) (c) \(4.92 \mathrm{~atm}\) (d) \(1.64 \mathrm{~atm}\)

Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Application of colligative properties are very useful in day-to-day life. One of its example is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles. A solution \(\mathrm{M}\) is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is \(0.9\) Given : Freezing point depression constant of water \(\left(K_{f}^{\text {water }}\right)\) $$ =1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} $$ Freezing point depression constant of ethanol ( \(\left.K_{f}{ }^{\text {ethanol }}\right)\) $$ =2.0 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} $$ Boiling point elevation constant of water \(\left(K_{b}^{\text {water }}\right)\) \(=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\) Boiling point elevation constant of ethanol \(\left(K_{b}^{\text {ethanol }}\right)=1.2 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\) Standard freezing point of water \(=273 \mathrm{~K}\) Standard freezing point of ethanol \(=155.7 \mathrm{~K}\) Standard boiling point of water \(=373 \mathrm{~K}\) Standard boiling point of ethanol \(=351.5 \mathrm{~K}\) Vapour pressure of pure water \(=32.8 \mathrm{~mm} \mathrm{Hg}\) Vapour pressure of pure ethanol \(=40 \mathrm{~mm} \mathrm{Hg}\) Molecular weight of water \(=18 \mathrm{~g} \mathrm{~mol}^{-1}\) Molecular weight of ethanol \(=46 \mathrm{~g} \mathrm{~mol}^{-1}\) In answering the following questions, consider the solution to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. The freezing point of the solution \(\mathrm{M}\) is (a) \(268.7 \mathrm{~K}\) (b) \(268.5 \mathrm{~K}\) (c) \(234.2 \mathrm{~K}\) (d) \(150.9 \mathrm{~K}\)

Determination of the molar mass of acetic acid in benzene using freezing point depression is affected by : (a) partial ionization (b) dissociation (c) complex formation (d) association

At room temperature, a dilute solution of urea is prepared by dissolving \(0.60 \mathrm{~g}\) of urea in \(360 \mathrm{~g}\) of water. If the vapour pressure of pure water at this temperature is \(35 \mathrm{mmHg}\), lowering of vapour pressure will be : (molar mass of urea \(=60 \mathrm{~g} \mathrm{~mol}^{-1}\) ) (a) \(0.027 \mathrm{mmHg}\) (b) \(0.028 \mathrm{mmHg}\) (c) \(0.017 \mathrm{mmHg}\) (d) \(0.031 \mathrm{mmHg}\)
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