Problem 24 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

4A quantity of \(10 \mathrm{~g}\) of solute 'A' and \(20 \mathrm{~g}\) of solute ' \(\mathrm{B}\) ' is dissolved in \(500 \mathrm{~m}\) l water. The solution is isotonic with the solution obtained by dissolving \(6.67 \mathrm{~g}\) of ' \(\mathrm{A}\) ' and \(30 \mathrm{~g}\) of ' \(\mathrm{B}\) ' in \(500 \mathrm{ml}\) water at the same temperature. The ratio of molar masses, \(M_{A}: M_{\mathrm{B}}\), is (a) \(1: 1\) (b) \(3: 1\) (c) \(1: 3\) (d) \(2: 3\)

Step-by-Step Solution

Answer

The ratio of the molar masses, \(M_A: M_B\), is 1:3.

Step 1: Understanding Isotonic Solutions

Two solutions are isotonic when they have the same osmotic pressure, which means that on a per-particle basis (moles per liter), they contain the same number of solute particles. In this context, it implies that the total moles of solute in the solutions are equal.

Step 2: Calculating Moles of Solute in First Solution

Let's denote the molar mass of solute A as \(M_A\) and the molar mass of solute B as \(M_B\). The number of moles of A in the first solution is \(\frac{10 \, \text{g}}{M_A}\), and the number of moles of B is \(\frac{20 \, \text{g}}{M_B}\). The total moles of solute in first solution is \(\frac{10}{M_A} + \frac{20}{M_B}\).

Step 3: Calculating Moles of Solute in Second Solution

Similarly, for the second solution, the number of moles of A is \(\frac{6.67 \, \text{g}}{M_A}\), and the number of moles of B is \(\frac{30 \, \text{g}}{M_B}\). The total moles of solute in second solution is \(\frac{6.67}{M_A} + \frac{30}{M_B}\).

Step 4: Equating Total Moles for Both Solutions

Since the solutions are isotonic, we can equate the total moles of the two solutions: \(\frac{10}{M_A} + \frac{20}{M_B} = \frac{6.67}{M_A} + \frac{30}{M_B}\).

Step 5: Solving the Equation

Subtracting the like terms on both sides gives us \(\frac{10 - 6.67}{M_A} = \frac{30 - 20}{M_B}\), which simplifies to \(\frac{3.33}{M_A} = \frac{10}{M_B}\). Now divide both sides by \(3.33\) to find the ratio \(\frac{M_A}{M_B}\), which is \(\frac{M_A}{M_B} = \frac{3.33}{10}\). This ratio simplifies further to \(\frac{1}{3}\) and thus, \(M_A:M_B = 1:3\).

Key Concepts

Osmotic Pressure

Imagine two solutions separated by a semi-permeable membrane which allows only the solvent (usually water) to pass through. Osmotic pressure is the pressure that needs to be applied to the solution to prevent the flow of water across the membrane. It's a measure of the tendency of water to move into the solution due to differences in solute concentration. In biological systems, maintaining osmotic balance is crucial for cell survival.

In the provided exercise, we talk about isotonic solutions, which means that the two solutions have the same osmotic pressure. This equilibrium prevents the net movement of water, ensuring that the cells in such an environment do not shrink or swell due to water loss or intake. Isotonic solutions are particularly important in medical settings— for instance, when intravenous (IV) fluids are administered to patients.

Molar Mass

Molar mass is the weight of one mole of a chemical element or a chemical compound, measured in grams per mole (g/mol). It corresponds to the mass of Avogadro's number (approximately 6.022 x 10²³) of molecules or atoms of the element or compound. The molar mass can be found on the periodic table for elements and calculated for compounds by summing the molar masses of the individual elements.

In our problem, molar masses of solutes 'A' and 'B' are significant for calculating the number of moles of each solute present in the solutions. Understanding the molar mass is crucial as it bridges the gap between the mass of a substance and the number of moles, and thereby, particles within a sample.

Calculating Moles

To quantify the amount of a substance, chemists use the unit 'mole'. Calculating moles provides a link between the mass of a substance and the number of particles contained. One mole is defined as the amount of substance that contains as many entities (atoms, molecules, ions or other particles) as there are in 12 grams of carbon-12.

In the given example, the moles of solute 'A' and 'B' are calculated using their masses and molar masses (mole = mass/molar mass). This calculation is foundational to solving the problem because the isotonic conditions imply that the total number of moles of the solutes should be the same in both solutions. Mastery in calculating moles is a must-have tool for students to analyze chemical reactions quantitatively.

Solution Concentration

The concentration of a solution refers to the amount of solute present in a given quantity of solvent or solution. One common way to express concentration is molarity, defined as the number of moles of solute per liter of solution (mol/L). In practical terms, it's a measure of how 'strong' or 'weak' a solution is.

When two solutions are isotonic, their concentrations, in terms of the number of solute particles per unit volume, are equal. Understanding solution concentration is critical in various fields, including chemistry, biology, and pharmacy. Control over the concentration allows scientists and medical professionals to predict how solutions will interact with cells, which is essential for creating safe and effective medical treatments.