Problem 79 - A Complete Resource Book in Chemistry for JEE Main

Question

The osmotic pressure at \(17^{\circ} \mathrm{C}\) of an aqueous solution containing \(1.75 \mathrm{~g}\) of sucrose per \(150 \mathrm{~mL}\) solution is (a) \(0.08 \mathrm{~atm}\) (b) \(8.1 \mathrm{~atm}\) (c) \(0.81\) (d) \(9.1 \mathrm{~atm}\)

Step-by-Step Solution

Answer

The osmotic pressure is (c) 0.81 atm.

Step 1: Understand the Formula

The osmotic pressure (\( \Pi \)) of a solution is given by the formula \( \Pi = iMRT \). Here, \(i\) is the van't Hoff factor (which is 1 for sucrose, a non-electrolyte), \(M\) 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.

Step 2: Calculate the Molarity

Molarity (\(M\)) is the number of moles of solute per liter of solution. Calculate the number of moles of sucrose (\(\text{C}_{12}\text{H}_{22}\text{O}_{11}\)) using its molecular weight of 342.3 g/mol. \[ \text{Moles of sucrose} = \frac{1.75 \, \text{g}}{342.3 \, \text{g/mol}} \approx 0.00511 \, \text{mol} \]Then, calculate molarity:\[ M = \frac{0.00511 \, \text{mol}}{0.150 \, \text{L}} \approx 0.03407 \, \text{mol/L} \]

Step 3: Convert Temperature to Kelvin

Convert the given temperature from Celsius to Kelvin by adding 273.15.\[ T = 17^{\circ} \text{C} + 273.15 = 290.15 \, \text{K} \]

Step 4: Plug Values into Osmotic Pressure Formula

Substitute the values into the osmotic pressure equation:\[ \Pi = (1)(0.03407 \, \text{mol/L})(0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1})(290.15 \, \text{K}) \]Calculate \(\Pi\):\[ \Pi \approx 0.812 \, \text{atm} \]

Step 5: Compare with Given Options

The calculated osmotic pressure, 0.812 atm, is closest to option (c) 0.81 atm. Therefore, the correct answer is (c) 0.81 atm.

Key Concepts

Van't Hoff Factor

The Van't Hoff factor (\( i \)) is a crucial parameter in calculating osmotic pressure, particularly when dealing with solutions. Simply put, it represents the number of particles into which a solute dissociates in solution. For instance, a solute that splits into three ions will have a Van't Hoff factor of 3.
For sucrose, a common non-electrolyte, the Van't Hoff factor is 1. This is because it does not dissociate into ions when dissolved in water. Instead, it remains as whole molecules, simplifying our osmotic pressure formula to \( \Pi = MRT \) when dealing with sucrose solutions.
Understanding this factor is essential:

• It helps in determining how solute particles behave in a solution.
• Influences colligative properties like boiling point elevation, freezing point depression, and of course, osmotic pressure.

The Van't Hoff factor provides insights into the nature of solutes and their interactions in solutions.

Molarity Calculation

Molarity (\( M \)) is a fundamental concept in chemistry that defines the concentration of a solution. It tells us how much solute is present in a given volume of solution, usually expressed in moles per liter (mol/L). To calculate molarity, follow these steps:
1. **Determine the moles of solute:** - Use the formula: \[ \text{Moles of solute} = \frac{\text{mass of solute (g)}}{\text{molar mass of solute (g/mol)}} \] - In our exercise, for sucrose: \[ \text{Moles of sucrose} = \frac{1.75 \, \text{g}}{342.3 \, \text{g/mol}} \approx 0.00511 \, \text{mol} \]2. **Calculate molarity:** - Next, factor in the volume of the solution in liters: \[ M = \frac{\text{moles of solute}}{\text{volume of solution (L)}} \] - For our case: \[ M = \frac{0.00511 \, \text{mol}}{0.150 \, \text{L}} \approx 0.03407 \, \text{mol/L} \]Understanding molarity helps:

• In preparing solutions of desired concentrations.
• In determining the stoichiometry of reactions in solution.

Determining molarity is a foundational skill for solving chemistry problems, including calculating osmotic pressure.

Temperature Conversion

Converting temperature from Celsius to Kelvin is a simple but vital step in many calculations in chemistry. This conversion is essential because Kelvin is the standard unit of temperature in scientific equations, including those for gas laws and colligative properties.
To convert:

• Add 273.15 to the Celsius temperature.
• This shifts the scale to Kelvin, which starts at absolute zero, the coldest possible temperature where all molecular motion stops.

For the given exercise, converting the temperature from 17°C to Kelvin:\[T = 17^{\circ}C + 273.15 = 290.15 \, \text{K} \]
This conversion ensures that all calculations, specifically those involving the ideal gas law or osmotic pressure (\( \Pi = iMRT \)), are done using a consistent and scientifically accepted scale.Understanding and applying temperature conversion is crucial when working with international science standards and ensuring accurate, comparable results across various equations and conditions.