Problem 91 - A Complete Resource Book in Chemistry for JEE Main

Question

\(20 \mathrm{~g}\) of a substance were dissolved in \(500 \mathrm{~mL}\) of water and the osmotic pressure of the solution was found to be \(600 \mathrm{~mm}\) of mercury at \(15^{\circ} \mathrm{C}\). The molecular weight of substance is (a) 998 (b) 1028 (c) 1098 (d) 1198

Step-by-Step Solution

Answer

The molecular weight of the substance is approximately 1198.

Step 1: Understand Osmotic Pressure Formula

The osmotic pressure (0"]) of a solution is given by the formula: \[\pi = \frac{n}{V}RT\]where \(n\) is the number of moles of solute, \(V\) is the volume of solution in liters, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin. We need to express \(n\) as \(\frac{w}{M}\), where \(w\) is the weight of the solute, and \(M\) is the molecular weight.

Step 2: Convert Units and Variables

We need to convert all the given quantities to the correct units. First, the osmotic pressure is given in millimeters of mercury, so we convert it to atmospheres (since 1 atm = 760 mmHg):\[\pi = \frac{600}{760} \text{ atm}\]The volume \(V\) is given as \(500 \text{ mL} = 0.5 \text{ L}\).The temperature \(T\) in Kelvin is found by adding 273.15 to the Celsius temperature:\[T = 15 + 273.15 = 288.15 \text{ K}\]The ideal gas constant \(R\) is \(0.0821 \text{ L atm/mol K}\).

Step 3: Rearrange Formula to Solve for Molecular Weight

Starting from the formula for osmotic pressure:\[\pi = \frac{w}{MV}RT \]we rearrange to solve for the molecular weight \(M\):\[M = \frac{wRT}{\pi V}\]where \(w = 20 \text{ g}\). Substitute the values into the rearranged equation.

Step 4: Substitute Values and Solve

Now, substitute the values into the equation:\[M = \frac{20 \times 0.0821 \times 288.15}{\left(\frac{600}{760}\right) \times 0.5}\]Calculate step-by-step:- Compute the osmotic pressure factor: \(\frac{600}{760} \approx 0.7895\)- Substitute all values:\[M = \frac{20 \times 0.0821 \times 288.15}{0.7895 \times 0.5}\]Calculate each part:- \(20 \times 0.0821 \times 288.15 \approx 473.49\)- \(0.7895 \times 0.5 = 0.39475\).So:\[M = \frac{473.49}{0.39475} \approx 1199\]

Step 5: Conclusion

After calculating, we find that the molecular weight \(M\) is approximately 1198, which matches option (d).

Key Concepts

Molecular Weight Calculation

Molecular weight, also known as molecular mass, is an important concept in chemistry. It represents the mass of one mole of a substance and is typically reported in grams per mole (g/mol). The molecular weight is derived from the sum of the atomic weights of all atoms in a molecule.

To calculate molecular weight using osmotic pressure, we use the formula from the colligative properties of solutions. The well-established formula for osmotic pressure is:

• \( \pi = \frac{w}{MV}RT \)
• Here, \( w \) is the weight of the solute (in grams), \( M \) is the molecular weight, \( V \) is the volume (in liters), \( R \) is the ideal gas constant, and \( T \) is the temperature (in Kelvin).

To solve for \( M \), the formula rearranges to:

• \[ M = \frac{wRT}{\pi V} \]

You can plug in the known values for \( w \), \( R \), \( T \), and \( \pi \) to find the unknown molecular weight \( M \). This method gives an accurate measurement crucial for identifying and studying different substances.

Ideal Gas Law

The Ideal Gas Law is a foundational principle in chemistry and physics. It describes how gases behave under various conditions of temperature, pressure, and volume. In this concept, the Ideal Gas Law formula is expressed as:

• \( PV = nRT \)
• \( P \) is the pressure (in atmospheres), \( V \) is the volume (in liters), \( n \) is the number of moles of the gas, \( R \) is the ideal gas constant, and \( T \) is the temperature (in Kelvin).

While the Ideal Gas Law is most commonly associated with gases, the principle also extends its application in calculating properties of solutions, such as osmotic pressure.

Osmotic pressure can simplify to the form of the Ideal Gas Law because it reflects the behavior of molecules when in a dilute solution, behaving similarly to gas particles in a container. For solutions, the simplification is:

• \( \pi = \frac{n}{V}RT \)

Therefore, understanding this concept is essential for solving problems involving solutions, pressures, and temperatures.

Solution Chemistry

Solution chemistry delves into how different substances dissolve in liquids to form solutions. A solution is a homogeneous mixture composed of two or more substances where the solute is uniformly distributed within the solvent.

There are some key properties about solutions that are crucial to understand:

• Solute and Solvent: The solute is the substance that is dissolved. The solvent is the substance in which the solute is dissolved, often in greater quantity.
• Concentration: This describes the amount of solute in a given volume of solvent or solution and can be expressed in several ways, such as molarity or molality.
• Colligative Properties: These properties depend on the number of solute particles in solution and not the identity of the solute. Osmotic pressure is one such colligative property and is crucial for calculating molecular weights in some cases.

Understanding how to manipulate and interpret these properties helps chemists in designing solutions with desired concentrations and characteristics, and also in analyzing experimental data from lab results.