A solution containing 8.6 g urea in one litre was found to be isotonic with 0.5% (mass/vol) solution of an organic, non volatile solute. The molecular mass of organic non volatile solute is:

Options

A. 348.9 g/mole
B. 34/89 g/mole
C. 3489 g/mole
D. 3.489 g/mole

Answer

To solve the problem, we need to find the molecular mass of the organic non-volatile solute that is isotonic with a solution containing 8.6 g of urea in 1 liter. We will use the concept of osmotic pressure and molarity.

### Step 1: Calculate the molarity of the urea solution
The formula for molarity (C) is given by:

\[ C = \frac{\text{mass of solute (g)}}{\text{molecular weight of solute (g/mol)} \times \text{volume of solution (L)}} \]

For urea (NH₂CONH₂), the molecular weight can be calculated as follows:

- Nitrogen (N): 14 g/mol × 2 = 28 g/mol
- Carbon (C): 12 g/mol × 1 = 12 g/mol
- Oxygen (O): 16 g/mol × 1 = 16 g/mol
- Hydrogen (H): 1 g/mol × 4 = 4 g/mol

Adding these together gives:

\[ \text{Molecular weight of urea} = 28 + 12 + 16 + 4 = 60 \text{ g/mol} \]

Now, substituting the values into the molarity formula:

\[ C_{\text{urea}} = \frac{8.6 \text{ g}}{60 \text{ g/mol} \times 1 \text{ L}} = \frac{8.6}{60} \approx 0.1433 \text{ mol/L} \]

### Step 2: Set up the equation for the isotonic solution
Since both solutions are isotonic, their molarities must be equal:

\[ C_{\text{urea}} = C_{\text{organic solute}} \]

Let \( M \) be the molecular weight of the organic non-volatile solute. The concentration of the organic solute in a 0.5% (mass/vol) solution can be calculated as follows:

A 0.5% solution means there are 0.5 g of solute in 100 mL of solution. To convert this to molarity:

\[ \text{Weight of solute} = 0.5 \text{ g} \]
\[ \text{Volume of solution} = 100 \text{ mL} = 0.1 \text{ L} \]

Now, the molarity \( C_{\text{organic solute}} \) is:

\[ C_{\text{organic solute}} = \frac{0.5 \text{ g}}{M \text{ g/mol} \times 0.1 \text{ L}} = \frac{5}{M} \text{ mol/L} \]

### Step 3: Equate the two molarities
Since the two solutions are isotonic:

\[ 0.1433 = \frac{5}{M} \]

### Step 4: Solve for M
Rearranging the equation gives:

\[ M = \frac{5}{0.1433} \approx 34.88 \text{ g/mol} \]

### Conclusion
The molecular mass of the organic non-volatile solute is approximately **34.88 g/mol**.

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