Osmotic pressure of a solution contaning 7 g.of a protein per `100 cm^(3)` of solution is `3.3 xx 10^(-2) ` atm at `37^(@)C` . Calculate the molar mass of protein.

Explanation

To calculate the molar mass of the protein from the given osmotic pressure, we can follow these steps:

### Step 1: Write the formula for osmotic pressure.
The formula for osmotic pressure (π) is given by:
\[ \pi = iCRT \]
where:
- \( \pi \) = osmotic pressure
- \( i \) = van 't Hoff factor (which is 1 for non-electrolytes like proteins)
- \( C \) = concentration of the solution in moles per liter (mol/L)
- \( R \) = universal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature in Kelvin

### Step 2: Convert the given temperature to Kelvin.
The temperature is given as 37°C. To convert this to Kelvin:
\[ T(K) = 37 + 273 = 310 \text{ K} \]

### Step 3: Calculate the concentration of the solution.
The concentration \( C \) can be calculated using the formula:
\[ C = \frac{\text{Number of moles of solute}}{\text{Volume of solution in liters}} \]
We know that:
- Mass of protein = 7 g
- Volume of solution = 100 cm³ = 0.1 L

Let \( M \) be the molar mass of the protein (in g/mol). The number of moles of protein is given by:
\[ \text{Number of moles} = \frac{\text{mass}}{M} = \frac{7}{M} \]
Thus, the concentration \( C \) becomes:
\[ C = \frac{7/M}{0.1} = \frac{70}{M} \text{ mol/L} \]

### Step 4: Substitute values into the osmotic pressure equation.
Using the values we have:
- \( \pi = 3.3 \times 10^{-2} \) atm
- \( i = 1 \)
- \( R = 0.0821 \) L·atm/(K·mol)
- \( T = 310 \) K

Substituting these into the osmotic pressure formula:
\[ 3.3 \times 10^{-2} = 1 \cdot \left(\frac{70}{M}\right) \cdot 0.0821 \cdot 310 \]

### Step 5: Solve for the molar mass \( M \).
Rearranging the equation to solve for \( M \):
\[ M = \frac{70 \cdot 0.0821 \cdot 310}{3.3 \times 10^{-2}} \]

Calculating the right-hand side:
1. Calculate \( 70 \cdot 0.0821 \cdot 310 \):
\[ 70 \cdot 0.0821 \cdot 310 \approx 70 \cdot 25.487 = 1784.09 \]
2. Now divide by \( 3.3 \times 10^{-2} \):
\[ M = \frac{1784.09}{0.033} \approx 53921.21 \text{ g/mol} \]

### Final Answer:
The molar mass of the protein is approximately \( 53921.21 \text{ g/mol} \).

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