Osomotic pressure of a solution containing 2 g dissolved protein per 300 `cm^3` of solution is 20 mm of Hg at `27^@C`. The molecular mass of protein is

Options

A. `"6239.6 g mol"^(-1)`
B. `"12315.5 g mol"^(-1)`
C. `"3692.1 g mol"^(-1)`
D. `"7368.4 g mol"^(-1)`

Explanation

To find the molecular mass of the protein using the given osmotic pressure, we will follow these steps:

### Step 1: Understand the formula for osmotic pressure
The osmotic pressure (π) can be expressed using the formula:
\[ \pi = C \cdot R \cdot T \]
Where:
- \( \pi \) = osmotic pressure
- \( C \) = concentration of the solution in moles per liter
- \( R \) = gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature in Kelvin

### Step 2: Convert the temperature to Kelvin
Given the temperature is \( 27^\circ C \):
\[ T = 27 + 273 = 300 \, K \]

### Step 3: Convert osmotic pressure from mm of Hg to atm
Given osmotic pressure is \( 20 \, mm \, of \, Hg \):
\[ \text{1 atm} = 760 \, mm \, of \, Hg \]
Thus,
\[ \pi = \frac{20}{760} \, atm = \frac{1}{38} \, atm \]

### Step 4: Calculate the concentration (C)
The concentration \( C \) can be expressed as:
\[ C = \frac{\text{number of moles of solute}}{\text{volume of solution in liters}} \]
The volume of the solution is given as \( 300 \, cm^3 \) which is \( 0.3 \, L \).

### Step 5: Express number of moles in terms of mass and molar mass
Let \( M \) be the molar mass of the protein. The number of moles of protein can be expressed as:
\[ \text{number of moles} = \frac{\text{mass}}{M} = \frac{2 \, g}{M} \]

### Step 6: Substitute values into the osmotic pressure formula
Substituting the values into the osmotic pressure formula:
\[ \frac{1}{38} = \left(\frac{2}{M \cdot 0.3}\right) \cdot 0.0821 \cdot 300 \]

### Step 7: Rearranging the equation to find M
Rearranging gives:
\[ \frac{1}{38} = \frac{2 \cdot 0.0821 \cdot 300}{M \cdot 0.3} \]
Now, solving for \( M \):
\[ M = \frac{2 \cdot 0.0821 \cdot 300 \cdot 38}{0.3} \]

### Step 8: Calculate the value of M
Calculating the right-hand side:
\[ M = \frac{2 \cdot 0.0821 \cdot 300 \cdot 38}{0.3} = \frac{2 \cdot 0.0821 \cdot 300 \cdot 38}{0.3} = 6239.6 \, g/mol \]

### Final Answer
Thus, the molecular mass of the protein is approximately:
\[ M \approx 6239.6 \, g/mol \]