10% (w/v) solution of sucrose (M.W = 342) is isotonic with 5% (w/v) solution of a solute X under identical conditions. The molecular weight of X is

Options

A. 171
B. 242
C. 684
D. 342

Answer

To find the molecular weight of solute X that is isotonic with a 10% (w/v) solution of sucrose, we can follow these steps:

### Step 1: Understand the Definitions
Isotonic solutions have the same osmotic pressure. For two solutions to be isotonic, their molarity must be equal.

### Step 2: Calculate the Molarity of the Sucrose Solution
- **Given**: 10% (w/v) solution of sucrose means there are 10 grams of sucrose in 100 mL of solution.
- **Molecular Weight of Sucrose (M.W)** = 342 g/mol.

**Calculate the number of moles of sucrose:**
\[ \text{Number of moles of sucrose} = \frac{\text{mass}}{\text{molecular weight}} = \frac{10 \text{ g}}{342 \text{ g/mol}} \approx 0.02924 \text{ mol} \]

**Convert the volume from mL to L:**
\[ 100 \text{ mL} = 0.1 \text{ L} \]

**Calculate the molarity of the sucrose solution:**
\[ \text{Molarity (M)} = \frac{\text{number of moles}}{\text{volume in L}} = \frac{0.02924 \text{ mol}}{0.1 \text{ L}} \approx 0.2924 \text{ M} \]

### Step 3: Calculate the Molarity of Solute X
- **Given**: 5% (w/v) solution of solute X means there are 5 grams of solute X in 100 mL of solution.

**Calculate the number of moles of solute X:**
Let the molecular weight of solute X be \( M_X \).

\[ \text{Number of moles of X} = \frac{5 \text{ g}}{M_X \text{ g/mol}} \]

**Calculate the molarity of the solute X solution:**
\[ \text{Molarity of X} = \frac{5 \text{ g}/M_X}{0.1 \text{ L}} = \frac{50}{M_X} \text{ M} \]

### Step 4: Set the Molarities Equal
Since the two solutions are isotonic, we can set their molarities equal to each other:
\[ 0.2924 \text{ M} = \frac{50}{M_X} \]

### Step 5: Solve for \( M_X \)
Rearranging the equation gives:
\[ M_X = \frac{50}{0.2924} \]

Calculating \( M_X \):
\[ M_X \approx 171 \text{ g/mol} \]

### Conclusion
The molecular weight of solute X is approximately **171 g/mol**.

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