Three grams of non - volatile solute whenn dissolved in a litre of water shows an osmotic pressure of 2 bar at 300K. Calculate the molar mass of the solute .`(R = 0.083 L bar K^(-1) mol^(-1)`) .

Explanation

To find the molar mass of the non-volatile solute, we can use the formula for osmotic pressure:

\[ \Pi = CRT \]

Where:
- \(\Pi\) is the osmotic pressure,
- \(C\) is the concentration (molarity) of the solution,
- \(R\) is the universal gas constant,
- \(T\) is the temperature in Kelvin.

### Step 1: Write down the given values
- Mass of solute (\(W_B\)) = 3 g
- Osmotic pressure (\(\Pi\)) = 2 bar
- Temperature (\(T\)) = 300 K
- Universal gas constant (\(R\)) = 0.083 L bar K\(^{-1}\) mol\(^{-1}\)
- Volume of solution (\(V\)) = 1 L

### Step 2: Rearrange the osmotic pressure formula to find molar mass
We know that:
\[ C = \frac{n}{V} \]
Where \(n\) is the number of moles of solute. The number of moles can be expressed as:
\[ n = \frac{W_B}{M_B} \]
Where \(M_B\) is the molar mass of the solute.

Substituting this into the concentration formula gives:
\[ C = \frac{W_B}{M_B \cdot V} \]

Now substituting \(C\) back into the osmotic pressure equation:
\[ \Pi = \left(\frac{W_B}{M_B \cdot V}\right)RT \]

### Step 3: Rearranging to find molar mass \(M_B\)
Rearranging the equation to solve for \(M_B\):
\[ M_B = \frac{W_B \cdot R \cdot T}{\Pi \cdot V} \]

### Step 4: Substitute the known values into the equation
Now, substituting the known values:
\[ M_B = \frac{3 \, \text{g} \cdot 0.083 \, \text{L bar K}^{-1} \text{mol}^{-1} \cdot 300 \, \text{K}}{2 \, \text{bar} \cdot 1 \, \text{L}} \]

### Step 5: Calculate the molar mass
Calculating the numerator:
\[ 3 \cdot 0.083 \cdot 300 = 74.7 \, \text{g bar K}^{-1} \text{mol}^{-1} \]

Now, calculating \(M_B\):
\[ M_B = \frac{74.7}{2} = 37.35 \, \text{g/mol} \]

### Final Answer
Thus, the molar mass of the solute is:
\[ \boxed{37.35 \, \text{g/mol}} \]