Step 1: Understanding the Concept:
Osmotic pressure (\(\pi\)) is defined as one of the four colligative properties of a solution. Colligative properties are unique because they depend solely on the ratio of the number of solute particles to the number of solvent molecules in a solution, rather than the chemical identity of the solute. In dilute solutions, the behavior of the solute mimics that of an ideal gas. Jacobus Henricus van 't Hoff established a mathematical relationship known as the van 't Hoff equation, which relates osmotic pressure to the molarity of the solution and the absolute temperature.
This property is exceptionally useful in analytical chemistry and biochemistry for determining the molar mass of complex molecules, especially polymers and proteins, because even small amounts of solute can produce a measurable osmotic pressure, whereas changes in boiling or freezing points might be too small to detect accurately.
Step 2: Key Formula or Approach:
The van 't Hoff equation for a non-electrolyte, non-volatile solute is:
\[ \pi = CRT \]
Where:
- \(\pi\) = Osmotic pressure (in atm)
- \(C\) = Molar concentration (moles per liter)
- \(R\) = Universal gas constant (0.0821 L$\cdot$atm/mol$\cdot$K)
- \(T\) = Absolute temperature (in Kelvin)
Since molarity \(C = \frac{n}{V}\) and moles \(n = \frac{w}{M}\), we can substitute these to find the molar mass \(M\):
\[ \pi = \frac{w}{M \cdot V} \cdot R \cdot T \]
Rearranging to solve for Molar Mass (\(M\)):
\[ M = \frac{w \cdot R \cdot T}{\pi \cdot V} \]
Step 3: Detailed Explanation:
Let us extract and convert all given data into appropriate units for calculation:
- Mass of solute (\(w\)) = 2 g
- Volume of solution (\(V\)) = 500 mL = 0.5 L
- Temperature in Celsius = 27$^{\circ}$C
- Temperature in Kelvin (\(T\)) = 27 + 273.15 $\approx$ 300 K
- Osmotic pressure (\(\pi\)) = 0.82 atm
- Gas constant (\(R\)) = 0.0821 L$\cdot$atm/mol$\cdot$K (For ease of calculation in exams, 0.082 is often used).
Now, substitute these values into the rearranged formula:
\[ M = \frac{2 \times 0.082 \times 300}{0.82 \times 0.5} \]
Simplifying the numerator:
\[ 2 \times 0.082 = 0.164 \]
\[ 0.164 \times 300 = 49.2 \]
Simplifying the denominator:
\[ 0.82 \times 0.5 = 0.41 \]
Performing the final division:
\[ M = \frac{49.2}{0.41} \]
Multiply both top and bottom by 100 to remove decimals:
\[ M = \frac{4920}{41} \]
By long division, \(4920 \div 41 = 120\).
Thus, the calculated molar mass of the unknown solute is exactly 120 g/mol. This procedure demonstrates how physical measurements of pressure and volume can lead directly to the molecular identity of a substance.
Step 4: Final Answer:
The molar mass of the solute is 120 g mol\(^{-1}\).
Therefore, the correct option is (2).