A solution of urea contains 8.6 g per litre. It is isotonic with 5% solution of a non-volatile solute. The molecular mass of the solute will be :

Options

A. 349
B. 34.9
C. 3490
D. 861

Answer

To solve the problem, we need to find the molecular mass of a non-volatile solute that makes a 5% solution isotonic with a solution of urea containing 8.6 g per liter. Here’s a step-by-step solution:

### Step 1: Calculate the number of moles of urea in the solution
Given:
- Mass of urea = 8.6 g
- Molecular mass of urea = 60 g/mol

Using the formula for moles:
\[ \text{Number of moles of urea} = \frac{\text{mass}}{\text{molecular mass}} = \frac{8.6 \, \text{g}}{60 \, \text{g/mol}} = 0.1433 \, \text{mol} \]

### Step 2: Calculate the osmotic pressure of the urea solution
The osmotic pressure (\(\Pi\)) is given by the formula:
\[ \Pi = CRT \]
Where:
- \(C\) = concentration in mol/L
- \(R\) = universal gas constant (0.0821 L·atm/(K·mol))
- \(T\) = temperature in Kelvin (assume room temperature, approximately 298 K)

Since we have 0.1433 moles in 1 L:
\[ \Pi_{\text{urea}} = 0.1433 \, \text{mol/L} \cdot R \cdot T \]

### Step 3: Calculate the concentration of the non-volatile solute
We know that the 5% solution of the non-volatile solute means there are 5 g of solute in 100 g of solution. Assuming the density of the solution is approximately 1 g/mL, the volume of the solution is:
\[ \text{Volume} = 100 \, \text{g} \approx 100 \, \text{mL} = 0.1 \, \text{L} \]

Now, the concentration of the non-volatile solute in moles is:
\[ C = \frac{\text{mass of solute}}{\text{molecular mass}} \cdot \frac{1}{\text{volume in L}} = \frac{5 \, \text{g}}{M} \cdot \frac{1}{0.1 \, \text{L}} = \frac{50}{M} \, \text{mol/L} \]

### Step 4: Set the osmotic pressures equal
Since the solutions are isotonic, their osmotic pressures are equal:
\[ \Pi_{\text{urea}} = \Pi_{\text{solute}} \]
Substituting the expressions for osmotic pressure:
\[ 0.1433 \cdot R \cdot T = \frac{50}{M} \cdot R \cdot T \]

### Step 5: Cancel out common terms
Since \(R\) and \(T\) are common on both sides, we can cancel them:
\[ 0.1433 = \frac{50}{M} \]

### Step 6: Solve for the molecular mass \(M\)
Rearranging gives:
\[ M = \frac{50}{0.1433} \approx 349.0 \, \text{g/mol} \]

### Conclusion
The molecular mass of the non-volatile solute is approximately 349 g/mol.