At 298 K, `1000 cm^(3)` of a solution cocontaining 4.34 g solute shows osmotic pressure of 2.55 atm. What is the molar mass of solute ? `(R=0.0821Latm K_(-1)mol^(-1))`

Options

A. `41.64 g mol^(-1)`
B. `82.73 g mol^(-1)`
C. `58.31 g mol^(-1)`
D. `91.65 g mol^(-1)`

Explanation

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

\[ \Pi = C \cdot R \cdot T \]

Where:
- \(\Pi\) = osmotic pressure (in atm)
- \(C\) = concentration (in moles per liter)
- \(R\) = ideal gas constant (0.0821 L·atm/(K·mol))
- \(T\) = temperature (in Kelvin)

### Step 1: Rearranging the formula
We can rearrange the formula to find the concentration \(C\):

\[ C = \frac{\Pi}{R \cdot T} \]

### Step 2: Substituting known values
Now, substitute the known values into the equation. Given:
- \(\Pi = 2.55 \, \text{atm}\)
- \(R = 0.0821 \, \text{L·atm/(K·mol)}\)
- \(T = 298 \, \text{K}\)

\[ C = \frac{2.55}{0.0821 \cdot 298} \]

### Step 3: Calculating concentration \(C\)
Now, calculate \(C\):

\[ C = \frac{2.55}{24.4758} \approx 0.1045 \, \text{mol/L} \]

### Step 4: Finding the number of moles
Next, we know that concentration \(C\) is also defined as:

\[ C = \frac{n}{V} \]

Where \(n\) is the number of moles and \(V\) is the volume in liters. Here, \(V = 1 \, \text{L}\) (since \(1000 \, \text{cm}^3 = 1 \, \text{L}\)).

So, rearranging gives:

\[ n = C \cdot V = 0.1045 \cdot 1 = 0.1045 \, \text{mol} \]

### Step 5: Calculating molar mass
Molar mass \(M\) can be calculated using the formula:

\[ M = \frac{m}{n} \]

Where:
- \(m = 4.34 \, \text{g}\) (mass of solute)
- \(n = 0.1045 \, \text{mol}\)

Substituting the values:

\[ M = \frac{4.34}{0.1045} \approx 41.5 \, \text{g/mol} \]

### Final Answer
Thus, the molar mass of the solute is approximately **41.5 g/mol**.

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