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 approximately 0.4 atm, matching option (c).

Step 1: Identify the Variables Required for Calculation

To find the osmotic pressure, we need to use the formula for osmotic pressure \( \pi = iCRT \). Here, \( i \) is the van 't Hoff factor (which we'll assume to be 1 since no dissociation is mentioned), \( C \) is the molarity of the solution, \( R \) is the universal gas constant \( 0.082 \, L \, atm \, K^{-1} \, mol^{-1} \), and \( T \) is the temperature in Kelvin.

Step 2: Convert Temperature to Kelvin

The temperature is given as \(27^{\circ}C\). We convert it to Kelvin by adding 273 to the Celsius temperature: \(27 + 273 = 300 \,K\).

Step 3: Calculate Molarity of the Solution

Molarity \( C \) is defined as the moles of solute per liter of solution. We have 4.0 g of solute and its molar mass is 246 g/mol, so moles of solute are \( \frac{4.0}{246} = 0.01626 \text{ mol/L} \). Thus, \( C = 0.01626 \, mol/L \).

Step 4: Calculate the Osmotic Pressure

Now substitute the values into the formula for osmotic pressure: \( \pi = CRT = 1 \times 0.01626 \times 0.082 \times 300 \). Calculate this to get the osmotic pressure \( \pi = 0.399 \approx 0.4 \, atm \).

Step 5: Determine the Most Suitable Option

The calculated osmotic pressure is approximately \(0.4 \, atm\), which matches option (c).

Key Concepts

Van 't Hoff Factor

One of the essential components in calculating osmotic pressure is understanding the Van 't Hoff factor, denoted as \( i \). It represents the number of particles a solute dissociates into when dissolved in a solution. In dilute solutions, for non-electrolytes like sugars, \( i \) is typically 1, as they don't dissociate. This contrasts with electrolytes, which may dissociate into multiple ions. For instance, a simple salt such as \( ext{NaCl} \) dissociates into two ions: \( ext{Na}^+ \) and \( ext{Cl}^- \), thus having an \( i \) value of 2.
For our specific problem, since no dissociation characteristics were mentioned, we presume the solute acts in a non-dissociative manner with \( i = 1 \). By using \( i \) accurately, you can determine how many effective particles are in a solution, and thus accurately calculate colligative properties like osmotic pressure. Considering \( i \) correctly is crucial when dealing with solutions of ionic compounds or when outcomes deviate from expected values.

Molarity Calculation

Molarity is a measure of how much solute is present in a given volume of solution. It's calculated by dividing the number of moles of solute by the volume of the solution in liters. Knowing the molarity allows you to understand the concentration of your solution, which is vital for calculating osmotic pressure.
In our example, the solute's weight is 4.0 grams with a molar mass of 246 g/mol. First, convert grams to moles with the equation:

• Moles of solute = \( \frac{4.0 \, \text{g}}{246 \, \text{g/mol}} \approx 0.01626 \, \text{mol} \)

Next, consider that the solution's volume is one liter, thus making the molarity (\( C \)):

• \( C = 0.01626 \, \text{mol/L} \)

This value provides the concentration needed for calculating osmotic pressure, highlighting how the molarity directly impacts the final pressure result under specific temperature and ideal solution assumptions.

Temperature Conversion to Kelvin

When calculating osmotic pressure, it's crucial to work with the temperature in Kelvin because the physical constants used in these expressions, such as the gas constant \( R \), are based on the Kelvin scale. The Kelvin temperature scale is absolute, meaning it starts at absolute zero, the point where molecular motion theoretically ceases.
To convert a Celsius temperature to Kelvin, add 273.15. However, for simplicity, many use simply 273 in calculations where high precision isn’t necessary. For example:

• Given temperature in Celsius: 27°C
• Converted to Kelvin: \( 27 + 273 = 300 \, K \)

Using Kelvin ensures compatibility with the other variables in equations like the ideal gas law and formulas for osmotic pressure, providing consistent and accurate results across scientific disciplines.