Problem 100 - A Complete Resource Book in Chemistry for JEE Main

Question

The osmotic pressure of a solution containing \(4.0\) \(\mathrm{g}\) of solute (molar mass 246 ) per litre at \(27^{\circ} \mathrm{C}\) is \(\left(\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)\) (a) \(0.1 \mathrm{~atm}\) (b) \(0.3 \mathrm{~atm}\) (c) \(0.4 \mathrm{~atm}\) (d) \(0.9 \mathrm{~atm}\)

Step-by-Step Solution

Answer

The osmotic pressure is 0.4 atm, matching option (c).

Step 1: Convert Temperature to Kelvin

First, we need to convert the given temperature from Celsius to Kelvin since the formula for osmotic pressure requires the temperature in Kelvin. The conversion formula is: \[ T(K) = T(C) + 273.15 \]For this problem: \[ T(K) = 27 + 273.15 = 300.15 \]We can round this to 300 K for simplicity.

Step 2: Calculate Molarity of the Solution

Molarity (M) is defined as the number of moles of solute per liter of solution. First, calculate the number of moles of the solute:\[ \text{moles of solute} = \frac{\text{mass of solute}}{\text{molar mass}} = \frac{4.0 \text{ g}}{246 \text{ g/mol}} \approx 0.0163 \text{ mol} \]Since the solution volume is 1 liter, the molarity is:\[ M = 0.0163 \text{ mol/L} \]

Step 3: Use Osmotic Pressure Formula

The osmotic pressure \( \Pi \) can be calculated using the formula:\[ \Pi = iMRT \]where:- \( i \) is the van 't Hoff factor (assumed to be 1 for a non-electrolyte),- \( M \) is the molarity determined above (0.0163 mol/L),- \( R \) is the gas constant 0.082 L atm K^{-1} mol^{-1},- \( T \) is the temperature in Kelvin (300 K).

Step 4: Calculate Osmotic Pressure

Substitute the known values into the osmotic pressure equation:\[ \Pi = (1)(0.0163)(0.082)(300) \]Calculate:\[ \Pi = 0.40026 \text{ atm} \]Rounding this value gives approximately 0.4 atm.

Step 5: Match with Multiple Choice

Compare the calculated osmotic pressure with the multiple-choice options given: (a) 0.1 atm (b) 0.3 atm (c) 0.4 atm (d) 0.9 atm The calculated osmotic pressure matches option (c) 0.4 atm.

Key Concepts

Molarity

Molarity is a way to express the concentration of a solution. It indicates how many moles of a solute are present in a certain volume of solution, usually measured in liters. In simple terms, it's a measure of how "strong" or "weak" a solution is based on the amount of substance dissolved in it.
To calculate molarity (M), you would use the formula:

• M = \( \frac{\text{moles of solute}}{\text{liters of solution}} \)

Let's consider the exercise where the solute has a mass of 4.0 g and a molar mass of 246 g/mol. First, determine the moles of solute by dividing the mass by the molar mass. Thus, the moles of solute are \( \frac{4.0 \text{ g}}{246 \text{ g/mol}} \approx 0.0163 \text{ mol} \).
If the solution volume is 1 liter, the molarity is simply 0.0163 mol/L. Knowing the molarity is crucial because it directly affects calculations involving osmotic pressure and other colligative properties.

van 't Hoff Factor

The van 't Hoff factor is an important concept when discussing solutions and their colligative properties. It represents the number of particles the solute yields in solution, which can affect properties like boiling point elevation, freezing point depression, and osmotic pressure.
For non-electrolytes, substances that do not ionize in solution, the van 't Hoff factor (i) is usually 1. This implies that the substance does not split into smaller ions or molecules but stays intact. However, for electrolytes (like salts), i can be greater than 1 since they dissociate into multiple ions upon dissolving.
In the case of the given exercise, since it’s assumed the solute does not dissociate, the van 't Hoff factor is 1. Thus, calculations for osmotic pressure use i = 1, simplifying the osmotic pressure formula to \( \Pi = iMRT = MRT \). It’s essential to understand the van 't Hoff factor to determine the correct impact of solutes on solution properties.

Temperature Conversion

Converting temperature is a fundamental step in thermodynamic calculations, such as those involving the osmotic pressure of solutions. Temperatures in scientific equations are often required in Kelvin, the SI unit for temperature.
To convert Celsius (°C) to Kelvin (K), use the formula:

• \( T(K) = T(°C) + 273.15 \)

This conversion is needed because Kelvin starts from absolute zero, providing a consistent baseline for scientific measurements.
In the example provided, the temperature was initially given in degrees Celsius as 27°C. By adding 273.15, we obtain a Kelvin temperature of 300.15 K. This is typically rounded to 300 K for simplicity in calculations. Always remember that using Kelvin is a must in osmotic pressure calculations, as it ensures accuracy and consistency in temperature-related equations.