Problem 102 - A Complete Resource Book in Chemistry for JEE Main

Question

An aqueous solution of sucrose \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\), containing \(34.2 \mathrm{~g} / \mathrm{L}\), has an osmotic pressure of \(2.38\) atmospheres at \(17^{\circ} \mathrm{C}\). For an aqueous solution of glucose \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) to be isotonic with this solution, it would have (a) \(18.0 \mathrm{~g} / \mathrm{L}\) (b) \(16.2 \mathrm{~g} / \mathrm{L}\) (c) \(36.6 \mathrm{~g} / \mathrm{L}\) of glucose (d) \(14.0 \mathrm{~g} / \mathrm{L}\)

Step-by-Step Solution

Answer

The glucose solution that is isotonic with the sucrose solution is 18.0 g/L, option (a).

Step 1: Understanding Osmotic Pressure Concept

Osmotic pressure (3939) is related to concentration (4747) of the solution through the equation 39 = 60 nRT/V). For a solution to be isotonic, it must have the same osmotic pressure as a reference solution.

Step 2: Determine the Molarity of Sucrose Solution

Given the osmotic pressure (2.38 atm) and temperature (17219C or 290 K), use the osmotic pressure formula 39 = 60 nRT/V) and solve for molarity, 4747, of sucrose. The R constant is 0.0821 L atm /(mol K). 39 = 2.38 atm = C 60 0.0821 60 290. Hence, C (sucrose) = 2.38 / (0.0821 60 290).

Step 3: Calculate Number of Moles per Liter for Sucrose

For sucrose, the molar weight = 342 g/mol. Using concentration from Step 2, multiply this molarity by the molar mass of sucrose to find g/L: 47 = 34.2 g/L; M = 47 / 342 g/mol to verify the concentration aligns with calculated molarity.

Step 4: Calculate Molarity of Glucose Solution

To determine the concentration of glucose needed to be isotonic, equate the molarities. As glucose (C6H12O6) requires the same molarity for isotonic conditions, use glucose's molar mass (180 g/mol) in the corresponding equation M (glucose) 2 60 g/L = 180 g/mol.

Step 5: Compare Calculated and Given Values for Glucose

You now know the required concentration for glucose. Calculate using glucose's molar mass and ensure it matches one of the provided options (a to d). Converting molecules to grams, M = C 60 molar mass glucose, and compare against 18.0, 16.2, 36.6, and 14.0 g/L.

Key Concepts

Molarity

Molarity is a way to express the concentration of a chemical solution. It is defined as the number of moles of solute per liter of solution. The formula for molarity is given by:M = \( \frac{n}{V} \) where:

• \( M \) is the molarity in moles per liter (mol/L)
• \( n \) is the number of moles of solute
• \( V \) is the volume of the solution in liters

In our problem, we needed to determine the molarity of the sucrose solution using its given osmotic pressure. By understanding molarity, students can comprehend the relative amount of solute present in a solution compared to its volume. To solve the original exercise, it was essential to calculate the molarity of sucrose based on the given conditions, and this was achieved using the formula for osmotic pressure and solving for molarity.

Isotonic Solutions

Isotonic solutions are solutions that have the same osmotic pressure. This means that when cells are placed in isotonic solutions, they neither gain nor lose water, maintaining cellular balance. Understanding isotonic solutions is crucial in the exercise to find the correct concentration of glucose that matches the given osmotic pressure of the sucrose solution. The goal was to find which concentration of glucose makes its solution isotonic with the sucrose solution. The osmotic pressure of a solution depends on the molarity and the temperature. For two solutions to be isotonic, they must have the same molarity when placed under the same conditions of temperature and pressure. This concept was used to calculate the molarity required for glucose so that its solution is isotonic with the sucrose solution. By equating the calculated molarity of the glucose solution to the potential given concentrations, one could determine the correct isotonic concentration.

Molecular Weight Calculations

Molecular weight calculations are vital for converting between mass and moles of a compound, allowing for the accurate determination of molarity. Molecular weight is the sum of the atomic weights of all the atoms in a molecule and is usually expressed in grams per mole (g/mol).To find the molarity in our exercise, knowing the molecular weights of sucrose (342 g/mol) and glucose (180 g/mol) was necessary. These values allow us to convert between grams per liter in the solution and moles per liter (molarity). By dividing the mass concentration (g/L) by the molecular weight of the compound, you obtain the molarity:\[ M = \frac{\text{mass per liter (g/L)}}{\text{molecular weight (g/mol)}} \] For sucrose and glucose, these weight calculations enabled us to determine their molarities, which were then used to compare their ability to create isotonic conditions.