A solution containing `8.6 g` urea in `1 L` was found to be isotonic with `5%(weight//volume)` solution of an organic non-volatile solute. The molecular weight of latter is

Options

A. 348.9
B. 34.89
C. 3489
D. 361.2

Explanation

To solve the problem, we need to find the molecular weight of the organic non-volatile solute that is isotonic with the urea solution. Let's break down the solution step by step.

### Step 1: Understand the Concept of Isotonic Solutions
Isotonic solutions have the same osmotic pressure. Therefore, the concentrations of the two solutions must be equal.

### Step 2: Calculate the Molarity of Urea Solution
Given:
- Mass of urea (w1) = 8.6 g
- Volume of solution (V1) = 1 L

The molar mass of urea (NH2CONH2) is approximately 60 g/mol.

First, we calculate the number of moles of urea:
\[ \text{Number of moles of urea} = \frac{\text{mass}}{\text{molar mass}} = \frac{8.6 \, \text{g}}{60 \, \text{g/mol}} = 0.1433 \, \text{mol} \]

Next, we calculate the molarity (C1) of the urea solution:
\[ C_1 = \frac{\text{Number of moles}}{\text{Volume in L}} = \frac{0.1433 \, \text{mol}}{1 \, \text{L}} = 0.1433 \, \text{mol/L} \]

### Step 3: Set Up the Equation for the Organic Solute
For the organic non-volatile solute, we have:
- Weight of solute (w2) = 5 g
- Volume of solution (V2) = 100 mL = 0.1 L
- Let the molecular weight of the organic solute be M2.

The molarity (C2) of the organic solute can be expressed as:
\[ C_2 = \frac{w_2}{M_2 \times V_2} \]
Substituting the known values:
\[ C_2 = \frac{5 \, \text{g}}{M_2 \times 0.1 \, \text{L}} = \frac{50}{M_2} \]

### Step 4: Set the Concentrations Equal
Since the solutions are isotonic, we set C1 equal to C2:
\[ 0.1433 = \frac{50}{M_2} \]

### Step 5: Solve for M2
Rearranging the equation gives:
\[ M_2 = \frac{50}{0.1433} \approx 348.9 \, \text{g/mol} \]

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

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