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 should have 18.0 g/L to be isotonic.

Step 1: Understanding Osmotic Pressure

Osmotic pressure is the pressure required to stop the flow of a solvent across a semipermeable membrane due to osmosis. The formula for osmotic pressure \( \Pi \) is given by \( \Pi = iCRT \), where \( i \) is the van 't Hoff factor, \( C \) is the molarity of the solution, \( R \) is the ideal gas constant \( (0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1}) \), and \( T \) is the temperature in Kelvin. For non-ionizing substances like sucrose and glucose, \( i = 1 \).

Step 2: Calculate Molarity of Sucrose Solution

First, find the molar mass of sucrose \( \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11} \):- \(12 \times 12.01 + 22 \times 1.008 + 11 \times 16.00 = 342.34 \, \text{g/mol}\).Given 34.2 g/L, the molarity \( C \) is given by: \[ C_\text{sucrose} = \frac{34.2}{342.34} = 0.0999 \, \text{mol/L} \]

Step 3: Calculate Temperature in Kelvin

Convert the temperature from Celsius to Kelvin:\[ T = 17 + 273.15 = 290.15 \, \text{K} \]

Step 4: Use Isotonic Property

Both the sucrose and glucose solutions should have the same osmotic pressure if isotonic. Equal osmotic pressures imply using the equation\[ \Pi = CRT \]this gives:\[ C_\text{sucrose}RT = C_\text{glucose}RT \] Thus, for isotonic solutions, the molarities must be equal: \( C_\text{sucrose} = C_\text{glucose} \).

Step 5: Calculate Molarity of Glucose Solution

Since the osmotic pressure must remain the same for isotonic conditions:\[ C_\text{glucose} = 0.0999 \, \text{mol/L} \]

Step 6: Calculate mass of Glucose for Isotonic Solution

The molar mass of glucose \( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \):- \(6 \times 12.01 + 12 \times 1.008 + 6 \times 16.00 = 180.18 \, \text{g/mol} \).To find the mass concentration:\[ \text{Mass} = 0.0999 \, \text{mol/L} \times 180.18 \, \text{g/mol} = 18.0 \, \text{g/L} \]

Step 7: Choose the Correct Option

Compare the calculated mass concentration with the given options: (a) 18.0 g/L (b) 16.2 g/L (c) 36.6 g/L (d) 14.0 g/L The correct answer is (a) 18.0 g/L.

Key Concepts

Isotonic Solutions

In the world of chemistry, isotonic solutions play an important role, especially in biological and medical contexts. When two solutions are termed "isotonic," they have the same osmotic pressure. This means that when placed on either side of a semipermeable membrane, there is no net movement of water molecules from one solution to the other.
This is crucial, for example, in medical treatments where intravenous fluids need to be isotonic with human blood. If not, it can lead to cells either swelling or shrinking due to water movement, negatively affecting cell function.
An isotonic solution is determined by its solute concentration. In the provided exercise, an aqueous solution needs to have the same osmotic pressure as another to be isotonic. This is accomplished by having equal molar concentrations as osmotic pressure is directly tied to molarity.

• The primary goal is balance — no net movement of water molecules.
• Equal concentrations mean equal osmotic pressures.

Understanding this ensures solutions can be used safely and effectively without disrupting cellular processes.

Molarity Calculations

Molarity is a key concept in chemistry that describes the concentration of a solute in a solution. It is expressed as moles of solute per liter of solution ( ext{mol/L}").
In the exercise, we calculated the molarity of sucrose and adjusted the glucose molarity to achieve isotonicity.
Molarity calculations help us ensure that solutions are prepared with precise concentrations, which is essential for chemical reactions and processes.
Here's a quick overview of the steps involved in finding molarity:

• Determine the mass of the solute (e.g., sucrose).
• Calculate the molar mass of the solute. For sucrose, it was 342.34 g/mol.
• Use the formula: \( C = \frac{\text{mass of solute}}{\text{molar mass}} \times \frac{1}{\text{volume of solution in L}} \).

Keeping these steps in mind allows for accurate calculations, ensuring that we achieve the desired properties in our solutions.

Semipermeable Membrane

A semipermeable membrane is a critical component in understanding osmotic processes. It is a material that allows certain molecules to pass through while blocking others. Typically, these membranes allow solvents (like water) to move while blocking solutes (like sucrose or glucose).
In biological systems, semipermeable membranes are key to maintaining cellular homeostasis. They control the internal environment by regulating the flow of substances. Osmosis, which occurs across these membranes, ensures that cells can efficiently uptake necessary nutrients and expel waste.

• Only solvents can move freely through, not larger solute particles.
• Important in contexts like dialysis and filtration.

Understanding semipermeable membranes is essential for comprehending how solutions interact biologically and chemically across barriers. This knowledge helps us manipulate conditions to benefit various applications, from medicine to chemistry.