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. `348.9 g mol^-1`
B. `174.5 g mol^-1`
C. `87.3g mol^-1`
D. `34.89 g mol^-1`

Answer

To solve the problem step-by-step, we will use the concept of osmotic pressure and the relationship between the concentration of solutes in isotonic solutions.

### Step 1: Understand the concept of isotonic solutions
Isotonic solutions have the same osmotic pressure. Therefore, we can equate the osmotic pressures of the two solutions:
\[ \pi_1 = \pi_2 \]
Where \(\pi_1\) is the osmotic pressure of the urea solution and \(\pi_2\) is the osmotic pressure of the non-volatile solute solution.

### Step 2: Write the formula for osmotic pressure
The osmotic pressure (\(\pi\)) is given by the formula:
\[ \pi = CRT \]
Where:
- \(C\) is the molarity of the solution,
- \(R\) is the universal gas constant,
- \(T\) is the temperature in Kelvin.

Since both solutions are isotonic, we can write:
\[ C_1RT = C_2RT \]
Here, \(C_1\) is the molarity of the urea solution and \(C_2\) is the molarity of the non-volatile solute solution.

### Step 3: Calculate the molarity of the urea solution
Given that the mass of urea is 8.6 g per liter, we need to calculate its molarity. The molecular weight of urea (NH₂CONH₂) is 60 g/mol.

\[ \text{Molarity (C)} = \frac{\text{mass (g)}}{\text{molecular weight (g/mol)} \times \text{volume (L)}} \]
Substituting the values:
\[ C_1 = \frac{8.6 \, \text{g}}{60 \, \text{g/mol} \times 1 \, \text{L}} = \frac{8.6}{60} \approx 0.1433 \, \text{mol/L} \]

### Step 4: Calculate the molarity of the non-volatile solute
We know that the 5% solution means there are 5 g of solute in 100 mL of solution. To find the molarity, we first convert 100 mL to liters:
\[ 100 \, \text{mL} = 0.1 \, \text{L} \]
Now we can express the molarity \(C_2\):
\[ C_2 = \frac{5 \, \text{g}}{M \times 0.1 \, \text{L}} = \frac{50}{M} \, \text{mol/L} \]
Where \(M\) is the molecular weight of the non-volatile solute.

### Step 5: Set the molarities equal
Since the two solutions are isotonic:
\[ C_1 = C_2 \]
Substituting the values we found:
\[ 0.1433 = \frac{50}{M} \]

### Step 6: Solve for the molecular weight \(M\)
Rearranging the equation to solve for \(M\):
\[ M = \frac{50}{0.1433} \approx 348.9 \, \text{g/mol} \]

### Conclusion
The molecular mass of the non-volatile solute is approximately **348.9 g/mol**.