The osmotic pressure of an aqueous solution containing 45g of sucrose `(343)` per litre of solution is `2.97` atm at `0^(@)C`. Find the value of the gas constant and compare the result with the accepted value.

Answer

To solve the problem, we will use the formula for osmotic pressure, which is given by:

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

Where:
- \(\Pi\) = osmotic pressure (in atm)
- \(C\) = concentration of the solution (in mol/L)
- \(R\) = gas constant (in L·atm/(K·mol))
- \(T\) = temperature (in Kelvin)

### Step 1: Calculate the concentration (C) of the sucrose solution

First, we need to find the number of moles of sucrose in the solution. The molar mass of sucrose is given as 342 g/mol.

\[ \text{Number of moles of sucrose} = \frac{\text{mass}}{\text{molar mass}} = \frac{45 \, \text{g}}{342 \, \text{g/mol}} \approx 0.131 \, \text{mol} \]

Since the solution volume is 1 L, the concentration \(C\) is:

\[ C = \frac{\text{Number of moles}}{\text{Volume in L}} = \frac{0.131 \, \text{mol}}{1 \, \text{L}} = 0.131 \, \text{mol/L} \]

### Step 2: Convert the temperature to Kelvin

The temperature given is \(0^\circ C\). To convert this to Kelvin:

\[ T = 0 + 273 = 273 \, \text{K} \]

### Step 3: Rearrange the osmotic pressure formula to find R

Now, we can rearrange the osmotic pressure formula to solve for \(R\):

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

### Step 4: Substitute the known values into the equation

Substituting the values we have:

\[ R = \frac{2.97 \, \text{atm}}{0.131 \, \text{mol/L} \cdot 273 \, \text{K}} \]

### Step 5: Calculate the value of R

Calculating the denominator:

\[ C \cdot T = 0.131 \, \text{mol/L} \cdot 273 \, \text{K} \approx 35.843 \, \text{mol·K/L} \]

Now substituting back into the equation for \(R\):

\[ R = \frac{2.97 \, \text{atm}}{35.843 \, \text{mol·K/L}} \approx 0.0828 \, \text{L·atm/(K·mol)} \]

### Step 6: Compare with the accepted value

The accepted value of the gas constant \(R\) is approximately \(0.0821 \, \text{L·atm/(K·mol)}\).

### Conclusion

The calculated value of \(R\) is \(0.0828 \, \text{L·atm/(K·mol)}\), which is very close to the accepted value of \(0.0821 \, \text{L·atm/(K·mol)}\).

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