1.46 g of a biopolymer dissolved in a 100 mL water at 300 K exerted an osmotic pressure of `2.42 xx 10^(-3)` bar.
The molar mass of the biopolymer is ________ ` xx 10^(4)g mol^(-1)`. (Round off to the Nearest Integer)

Answer

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

\[ \Pi = iCRT \]

Where:
- \(\Pi\) = osmotic pressure
- \(i\) = van 't Hoff factor (which is 1 for non-electrolytes)
- \(C\) = molar concentration of the solution
- \(R\) = universal gas constant
- \(T\) = temperature in Kelvin

### Step 1: Identify the given values
- Mass of the biopolymer (solute), \(m = 1.46 \, \text{g}\)
- Volume of the solution, \(V = 100 \, \text{mL} = 0.1 \, \text{L}\)
- Osmotic pressure, \(\Pi = 2.42 \times 10^{-3} \, \text{bar}\)
- Temperature, \(T = 300 \, \text{K}\)
- Gas constant, \(R = 0.083 \, \text{bar L K}^{-1} \text{mol}^{-1}\)

### Step 2: Calculate the concentration \(C\)
Since the osmotic pressure formula can be rearranged to find concentration, we have:

\[ C = \frac{\Pi}{iRT} \]

Substituting \(i = 1\):

\[ C = \frac{\Pi}{RT} = \frac{2.42 \times 10^{-3} \, \text{bar}}{0.083 \, \text{bar L K}^{-1} \text{mol}^{-1} \times 300 \, \text{K}} \]

Calculating the denominator:

\[ 0.083 \times 300 = 24.9 \, \text{bar L mol}^{-1} \]

Now substituting back into the concentration formula:

\[ C = \frac{2.42 \times 10^{-3}}{24.9} \approx 9.73 \times 10^{-5} \, \text{mol L}^{-1} \]

### Step 3: Calculate the number of moles \(n\)
The number of moles \(n\) can be calculated using the concentration and volume:

\[ n = C \times V = (9.73 \times 10^{-5} \, \text{mol L}^{-1}) \times (0.1 \, \text{L}) = 9.73 \times 10^{-6} \, \text{mol} \]

### Step 4: Calculate the molar mass \(M\)
The molar mass \(M\) is given by:

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

Substituting the values:

\[ M = \frac{1.46 \, \text{g}}{9.73 \times 10^{-6} \, \text{mol}} \approx 150,000 \, \text{g mol}^{-1} \]

### Step 5: Convert to the required format
To express this in the format of \(x \times 10^4 \, \text{g mol}^{-1}\):

\[ M \approx 15 \times 10^4 \, \text{g mol}^{-1} \]

### Step 6: Round off to the nearest integer
The nearest integer is \(15\).

### Final Answer
The molar mass of the biopolymer is \(15 \times 10^4 \, \text{g mol}^{-1}\).

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