Problem 122 - A Complete Resource Book in Chemistry for JEE Main

Question

The strength of \(0.01 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\) solution in terms of molality is nearly ______ when the density of the solution is \(1.1 \mathrm{~g} / \mathrm{mL}\) (a) \(9 \times 10^{-3}\) (b) \(1.8 \times 10^{-3}\) (c) \(4.5 \times 10^{-3}\) (d) \(1.1 \times 10^{-3}\)

Step-by-Step Solution

Answer

The molality of the solution is approximately \(9 \times 10^{-3}\) m, option (a).

Step 1: Understanding Molality

Molality is defined as the number of moles of solute per kilogram of solvent. The formula is given by \(m = \frac{n}{M_s}\), where \(n\) is the moles of solute and \(M_s\) is the mass of the solvent in kilograms.

Step 2: Calculate Moles of Solute

The molarity of the solution is given as 0.01 M, which means there are 0.01 moles of \(\mathrm{Na}_2\mathrm{CO}_3\) in 1 liter of solution. Therefore, the number of moles (\(n\)) of \(\mathrm{Na}_2\mathrm{CO}_3\) per liter is 0.01.

Step 3: Determine Mass of Solution

The density of the solution is given as \(1.1 \, \mathrm{g/mL}\). Therefore, the mass of 1 liter (1000 mL) of solution is \(1.1 \, \mathrm{g/mL} \times 1000 \, \mathrm{mL} = 1100 \, \mathrm{g}\).

Step 4: Calculate Mass of Solvent

The mass of the solvent \(M_s\) can be found by subtracting the mass of the solute from the total mass of the solution. The mass of \(0.01\) moles of \(\mathrm{Na}_2\mathrm{CO}_3\), with a molar mass of around \(106 \, \mathrm{g/mol}\), is \(0.01 \, \mathrm{mol} \times 106 \, \mathrm{g/mol} = 1.06 \, \mathrm{g}\). Thus, the mass of the solvent is \(1100 \, \mathrm{g} - 1.06 \, \mathrm{g} = 1098.94 \, \mathrm{g}\) or \(1.09894 \, \mathrm{kg}\).

Step 5: Calculate Molality

Using the formula for molality \(m = \frac{n}{M_s}\), where \(n = 0.01\, \mathrm{mol}\) and \(M_s = 1.09894 \, \mathrm{kg}\), we have:\[m = \frac{0.01}{1.09894} \approx 0.0091 \approx 9 \times 10^{-3} \mathrm{~m}\]

Step 6: Choose the Correct Answer

The calculated molality \(9 \times 10^{-3}\) matches option (a). Therefore, the correct answer is (a) \(9 \times 10^{-3}\).

Key Concepts

Molarity

Molarity is an essential concept in solution chemistry, which helps you understand how concentrated a solution is. Specifically, it describes the number of moles of solute present in one liter of solution. The formula to calculate molarity is: \[M = \frac{n}{V} \]where

• \(M\) is the molarity,
• \(n\) is the number of moles of solute,
• \(V\) is the volume of the solution in liters.

Knowing the molarity helps in various applications, such as diluting solutions or mixing reactants in stoichiometric reactions. In the depicted problem, the molarity of the \(\mathrm{Na}_2\mathrm{CO}_3\) solution is given as \(0.01\) M, indicating that every liter of this solution contains \(0.01\) moles of \(\mathrm{Na}_2\mathrm{CO}_3\).

Density

Density is another important property in chemistry, providing insight into the mass per unit volume of a substance or solution. It is typically expressed in g/mL or kg/L. The formula is:\[\text{Density} = \frac{\text{Mass}}{\text{Volume}}\]In the context of solution chemistry, knowing the density allows you to calculate the mass of the solution when given the volume. For instance, if the solution has a density of \(1.1 \, \mathrm{g/mL}\), as stated in the exercise, you can find that 1 liter (1000 mL) of it has a mass of \(1.1 \times 1000 = 1100 \; \mathrm{g}\). This relationship aids in converting between the volume and mass during the calculation of other properties, such as molality.

Solution Chemistry

Solution chemistry revolves around the interaction of solutes and solvents in forming solutions. A solution consists of a solute, which is the substance that is dissolved, and a solvent, which is the medium in which dissolution occurs. Understanding the interplay between components in a solution involves concepts of concentration, such as molarity and molality, to describe how much solute is present relative to the solvent.

Role of Solvent and Solute

- Solvent: In this exercise, the water acts as the solvent, which dissolves \(\mathrm{Na}_2\mathrm{CO}_3\) to create the solution.- Solute: \(\mathrm{Na}_2\mathrm{CO}_3\) is the solute, the substance whose concentration we are interested.In essence, the study of solution chemistry helps in predicting how solutions will behave in reactions and how changes in conditions, like temperature or pressure, will affect their concentrations.

Stoichiometry

Stoichiometry is central to balancing chemical equations and making predictions about the outcomes of chemical reactions. It centers around the quantitative relationships between the amounts of reactants and products. In the calculation of molality, understanding stoichiometry helps evaluate the mass of the solvent and solute in the solution.

Practical Application

- For the given problem, the stoichiometry principles help determine that the molar mass of \(\mathrm{Na}_2\mathrm{CO}_3\) is approximately \(106 \, \mathrm{g/mol}\).- By determining how many grams of \(\mathrm{Na}_2\mathrm{CO}_3\) are present per \(0.01\) moles, you can calculate the mass of the solvent and subsequently determine molality.

Importance in Calculations

- Stoichiometry enables the step-by-step calculation where knowing the moles of solute and total solution mass leads to the calculation of solvent mass. - Consequently, this facilitates conversion into molality, as shown in the original exercise's steps. Breaking down these tasks makes complex problems in chemistry manageable, especially when determining concentrations and reacting masses.