Problem 18 - A Complete Resource Book in Chemistry for JEE Main

Question

How many grams of \(\mathrm{KCl}\) must be added to \(75 \mathrm{~g}\) of water to produce a solution with a molality of \(2.25\) (a) \(1.257 \mathrm{~g}\) (b) \(125.7 \mathrm{~g}\) (c) \(12.57 \mathrm{~g}\) (d) \(25.14 \mathrm{~g}\)

Step-by-Step Solution

Answer

The correct answer is (c) 12.57 g.

Step 1: Understanding Molality

Molality (\(m\)) is defined as the number of moles of solute per kilogram of solvent. It is given by the equation:\[m = \frac{\text{moles of solute}}{\text{kg of solvent}}.\]

Step 2: Identify Given Values and Required Conversion

We are given that the molality \(m\) is \(2.25\) and the mass of the solvent (water) is \(75 \text{ g}\) = \(0.075 \text{ kg}\). We need to find the moles of \(\mathrm{KCl}\) using these values.

Step 3: Calculate Moles of KCl Needed

Using the equation for molality, we can calculate the moles of \(\mathrm{KCl}\) needed:\[2.25 = \frac{\text{moles of } \mathrm{KCl}}{0.075 \text{ kg}}\]Rearranging to solve for the moles of KCl:\[\text{moles of } \mathrm{KCl} = 2.25 \times 0.075 = 0.16875\text{ moles}\]

Step 4: Determine Molar Mass of KCl

The molar mass of \(\mathrm{KCl}\) is the sum of the molar masses of potassium (K) and chlorine (Cl). From the periodic table, the molar mass of potassium is approximately \(39.1 \text{ g/mol}\) and chlorine is \(35.5 \text{ g/mol}\). Hence, the molar mass of \(\mathrm{KCl}\) is:\[39.1 + 35.5 = 74.6 \text{ g/mol}\]

Step 5: Calculate Grams of KCl Needed

Convert moles of \(\mathrm{KCl}\) to grams using the molar mass:\[\text{grams of } \mathrm{KCl} = 0.16875 \text{ moles} \times 74.6 \text{ g/mol} = 12.583 \text{ g}\]Rounding to the proper significant figures, we find it requires approximately \(12.57 \text{ g}\) of \(\mathrm{KCl}\).

Step 6: Choose the Correct Answer

From the options given: (a) 1.257 g, (b) 125.7 g, (c) 12.57 g, (d) 25.14 g, the correct answer is (c) 12.57 g.

Key Concepts

Moles of Solute

Understanding the concept of moles is key when dealing with solutions like the one given in the exercise. In a solution, the solute is the substance that is dissolved. To find the number of moles of a solute, such as potassium chloride (\(KCl\)), we can use the formula:

• \( \text{moles of solute} = \frac{\text{mass of solute (g)}}{\text{molar mass (g/mol)}} \)

By knowing the molality, which is the concentration expressed as moles per kilogram of solvent, and the weight of the solvent, we can calculate the required moles of the solute.
Molality, symbolized as \(m\), is especially useful in these calculations because it does not change with temperature fluctuations unlike molarity.
The exercise provides a target molality of \(2.25\). We rearrange the formula to find the moles of \(KCl\), demonstrating this essential use of moles in chemistry. Multiply \(2.25 \text{ mol/kg} \) by the solvent weight (\(0.075 \text{ kg}\)), to obtain \(0.16875\) moles of \(KCl\) that are needed.

Molar Mass

Molar mass is the mass of one mole of a given substance and is expressed in grams per mole (g/mol). It reflects the combined weight of all atoms in a molecule. In our case, understanding the molar mass of potassium chloride (\(KCl\)) is crucial for converting moles to grams.

To calculate the molar mass of \(KCl\), consider the atomic masses of its constituent elements:

• Potassium (K): approximately \(39.1 \text{ g/mol}\)
• Chlorine (Cl): approximately \(35.5 \text{ g/mol}\)

The sum of these atomic masses gives us the molar mass of \(KCl\): \[\text{molar mass of } KCl = 39.1 + 35.5 = 74.6 \text{ g/mol}\]This calculation is necessary to convert \(0.16875\) moles of \(KCl\) to grams, which is effectively done by multiplying the moles by the molar mass \(0.16875 \text{ moles} \times 74.6 \text{ g/mol} = 12.583 \text{ grams of } KCl\).

Potassium Chloride

Potassium chloride, which is abbreviated as \(KCl\), is commonly used in chemistry as a source of potassium ions. It is a white crystalline solid that is soluble in water, making it useful for solutions.

In the context of the problem, potassium chloride is the solute that needs to be dissolved to achieve a specific molality.
The task in the exercise is to convert a certain number of moles, already determined from the given molality and solvent weight, into grams for practical measurement and use in solution preparation.

• \(KCl\) is widely used in industrial processes, as a de-icing agent, and even in medical treatments for electrolyte imbalances.
• It is important to precisely weigh \(12.57 \text{ g}\) of potassium chloride to meet the molality condition of the solution described in the task.

This exercise showcases the practical application of molecular chemistry concepts to real-world problems.