A `5%` solution of cane sugar (molecular weight=342) is isotonic with `1%` solution of substance `X`.The molecular weight of `X` is

Options

A. 171.2
B. 68.4
C. 34.2
D. 136.2

Explanation

To solve the problem, we need to determine the molecular weight of substance X based on the information provided about the isotonic solutions of cane sugar and substance X.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- A 5% solution of cane sugar means there are 5 grams of cane sugar in 100 mL of solution.
- The molecular weight of cane sugar is given as 342 g/mol.
- A 1% solution of substance X means there is 1 gram of substance X in 100 mL of solution.

2. **Setting Up the Problem**:
- Since the two solutions are isotonic, their osmotic pressures are equal. This leads us to set the concentrations equal to each other.
- The formula for osmotic pressure (π) is given by:
\[ \pi = C \cdot R \cdot T \]
- For isotonic solutions, we can say:
\[ C_1 = C_2 \]
- Where \(C_1\) is the concentration of cane sugar and \(C_2\) is the concentration of substance X.

3. **Calculating the Concentration of Cane Sugar**:
- The concentration \(C_1\) can be calculated using the formula:
\[ C_1 = \frac{\text{Number of moles of solute}}{\text{Volume of solution in liters}} \]
- The number of moles of cane sugar:
\[ \text{Moles of cane sugar} = \frac{W_1}{M_1} = \frac{5 \text{ g}}{342 \text{ g/mol}} \approx 0.0146 \text{ moles} \]
- The volume of solution in liters is:
\[ V_1 = 100 \text{ mL} = 0.1 \text{ L} \]
- Therefore, the concentration \(C_1\) is:
\[ C_1 = \frac{0.0146 \text{ moles}}{0.1 \text{ L}} = 0.146 \text{ moles/L} \]

4. **Calculating the Concentration of Substance X**:
- For substance X, we have:
\[ C_2 = \frac{W_2/M_2}{V_2} \]
- Here, \(W_2 = 1 \text{ g}\) and \(V_2 = 100 \text{ mL} = 0.1 \text{ L}\).
- Thus, we can express the concentration \(C_2\) as:
\[ C_2 = \frac{1 \text{ g}/M_2}{0.1 \text{ L}} = \frac{10}{M_2} \text{ moles/L} \]

5. **Setting the Concentrations Equal**:
- Since \(C_1 = C_2\):
\[ 0.146 = \frac{10}{M_2} \]
- Rearranging gives:
\[ M_2 = \frac{10}{0.146} \approx 68.49 \text{ g/mol} \]

6. **Final Answer**:
- The molecular weight of substance X is approximately **68.4 g/mol**.