Question

A solution is prepared by dissolving \(2\) gram of non-volatile solute in \(500\) mL solution at \(27^\circ\text{C}\). The osmotic pressure of the solution is \(0.82\) atm. The molar mass of the solute is:

• A. \(100\ \text{g mol}^{-1}\)
• B. \(120\ \text{g mol}^{-1}\)
• C. \(150\ \text{g mol}^{-1}\)
• D. \(180\ \text{g mol}^{-1}\)

Hint:

For osmotic pressure problems, always convert:
• Temperature into Kelvin
• Volume into liters A commonly used formula is: \[ M = \frac{wRT}{\pi V} \] which directly gives molar mass.

Answer 1

The Correct Option is A

Concept: Osmotic pressure is a colligative property that depends on the number of solute particles present in the solution. For dilute solutions, osmotic pressure is given by the van't Hoff equation: \[ \pi = CRT \] where:
• \(\pi\) = osmotic pressure
• \(C\) = molar concentration of solution
• \(R\) = universal gas constant
• \(T\) = absolute temperature Since: \[ C = \frac{n}{V} = \frac{w}{MV} \] the osmotic pressure equation becomes: \[ \pi = \frac{wRT}{MV} \] where:
• \(w\) = mass of solute
• \(M\) = molar mass of solute
• \(V\) = volume of solution in liters

Step 1: Writing the given data.
From the question: \[ w = 2\ \text{g} \] \[ V = 500\ \text{mL} = 0.5\ \text{L} \] \[ T = 27^\circ\text{C} = 27 + 273 = 300\ \text{K} \] \[ \pi = 0.82\ \text{atm} \] Gas constant: \[ R = 0.082\ \text{L atm mol}^{-1}\text{K}^{-1} \]

Step 2: Substituting values into osmotic pressure formula.
Using: \[ \pi = \frac{wRT}{MV} \] Substituting all values: \[ 0.82 = \frac{2 \times 0.082 \times 300}{M \times 0.5} \]

Step 3: Simplifying the numerator.
\[ 2 \times 0.082 \times 300 \] \[ = 49.2 \] Therefore: \[ 0.82 = \frac{49.2}{0.5M} \]

Step 4: Solving for molar mass \(M\).
Multiply both sides by \(0.5M\): \[ 0.82 \times 0.5M = 49.2 \] \[ 0.41M = 49.2 \] Now divide both sides by \(0.41\): \[ M = \frac{49.2}{0.41} \] \[ M = 120\ \text{g mol}^{-1} \]

Step 5: Identifying the correct option.
The molar mass of the non-volatile solute is: \[ \boxed{120\ \text{g mol}^{-1}} \] Hence, the correct answer is: \[ \boxed{(2)\ 120\ \text{g mol}^{-1}} \]