Problem 8 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

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) \(1.51\)

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: Identify Given Values

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

Step 3: 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.

Step 4: Convert Solvent Mass to Kilograms

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

Step 5: Calculate Moles of KI

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

Step 6: Compute Molality

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

Step 7: Select the Correct Answer

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

Key Concepts

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.