Question

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

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

Hint:

Always convert temperature to Kelvin and volume to liters before applying colligative property formulas. Also try cancelling constants like \(R = 0.082\) with given values like \(0.82\) to simplify calculations quickly.

Answer 1

The Correct Option is B

Concept: Osmotic pressure (\(\pi\)) is a colligative property that depends on the number of solute particles in solution. For a dilute solution of a non-volatile, non-electrolyte solute, it is given by the van’t Hoff equation: \[ \pi = CRT \] where \(C\) is molar concentration, \(R\) is the gas constant, and \(T\) is absolute temperature. Since \(C = \frac{w}{MV}\), the formula becomes: \[ \pi = \frac{wRT}{MV} \] where \(w\) is mass of solute and \(M\) is molar mass.

Step 1: Given data and unit conversion.

• Mass of solute, \(w = 2\,\text{g}\)
• Volume of solution, \(V = 500\,\text{mL} = 0.5\,\text{L}\)
• Temperature, \(T = 27^{\circ}\text{C} = 300\,\text{K}\)
• Osmotic pressure, \(\pi = 0.82\,\text{atm}\)
• Gas constant, \(R = 0.082\,\text{L atm K}^{-1}\text{ mol}^{-1}\)

Step 2: Rearranging formula for molar mass.
\[ M = \frac{wRT}{\pi V} \]

Step 3: Substituting values.
\[ M = \frac{2 \times 0.082 \times 300}{0.82 \times 0.5} \] \[ M = \frac{2 \times 24.6}{0.41} \] Now simplify in a smarter way: \[ M = \left(\frac{0.082}{0.82}\right)\left(\frac{2 \times 300}{0.5}\right) \] \[ M = \left(\frac{1}{10}\right)(1200) \] \[ M = 120\,\text{g mol}^{-1} \]