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 isotonic glucose solution has a concentration of (a) 18.0 g/L.

Step 1: Understanding Osmotic Pressure Formula

To find an isotonic solution, we need to understand the formula for osmotic pressure: \( \pi = iCRT \), where \( \pi \) is the osmotic pressure, \( i \) is the van't Hoff factor, which is 1 for both sucrose and glucose, \( C \) is the molarity, \( R \) is the ideal gas constant \(0.0821\, \text{L atm K}^{-1} \text{mol}^{-1}\), and \( T \) is the temperature in Kelvin.

Step 2: Calculate Molarity of Sucrose

We already know \( \pi = 2.38 \text{ atm} \), \( T = 17 + 273 = 290 \text{ K} \). Rearrange the formula to find \( C \): \( C = \frac{\pi}{iRT} = \frac{2.38}{1 \times 0.0821 \times 290} \approx 0.0997 \text{ mol/L} \).

Step 3: Calculate Molarity of Glucose

For the glucose solution to be isotonic, the molarity must be the same. Therefore, for glucose, \( C_{\text{glucose}} = 0.0997 \text{ mol/L} \).

Step 4: Convert Molarity of Glucose to Concentration

We need to find the mass of glucose in grams per liter. The molar mass of glucose \( C_6H_{12}O_6 \) is \(12 \times 6 + 1 \times 12 + 16 \times 6 = 180 \text{ g/mol} \). Using \( C_{\text{glucose}} \), the mass is \( 0.0997 \text{ mol/L} \times 180 \text{ g/mol} \approx 17.946 \text{ g/L} \).

Step 5: Select the Correct Option

The calculated mass of glucose needed is approximately \( 17.946 \text{ g/L} \), which is closest to \( 18.0 \text{ g/L} \) in the given options.

Key Concepts

Isotonic Solutions

The concept of isotonic solutions is pivotal in understanding the behavior of solutions when separated by a semi-permeable membrane. An isotonic solution, essentially, has the same osmotic pressure as another solution when at equilibrium. This means that the solute concentration—and hence, the number of solute particles—in both solutions is equal. When two solutions are isotonic, there is no net movement of water across the membrane, preventing cell size from changing, which is crucial in many biological contexts. In the exercise, the isotonic condition is achieved when the glucose solution has the same osmotic pressure as the sucrose solution. This ensures the two solutions can be compared for practical applications like medical treatments, where isotonic solutions often prevent cell damage.

Molarity Calculation

Molarity is a concentration unit that expresses moles of solute per liter of solution (mol/L). It is a key concept for calculating osmotic pressure in solutions. To determine the molarity from osmotic pressure, you can use the formula:

• \( C = \frac{\pi}{iRT} \)

where \( \pi \) is the osmotic pressure, \( i \) is the van't Hoff factor, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin. In the problem, the molarity needed for both sucrose and glucose solutions is calculated by rearranging the equation to solve for \( C \). For sucrose, this calculation helps in understanding how much solute is present per liter and sets the stage for calculating how much glucose is needed to be isotonic with the sucrose solution.

Van't Hoff Factor

The van't Hoff factor, \( i \), is essential in determining the effect of a solute on the osmotic pressure of a solution. It represents the number of particles a solute splits into or forms in a solution. For simple sugars like sucrose and glucose, \( i \) is 1 because they do not dissociate into ions in solution. In more complex ionic solutions, \( i \) might be greater than 1, reflecting more particles than initially thought. This factor is included in the osmotic pressure formula \( \pi = iCRT \), which helps in accurately describing the concentration of particles affecting the pressure experienced by the solution. Being aware of \( i \) is crucial for scientists and engineers to precisely gauge solution behavior in various fields, from chemistry to biology and environmental science.