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

Options

A. 342
B. 171.12
C. 65.6
D. 136.8

Explanation

To solve the problem, we need to find the molecular weight of substance X, given that a 5% solution of cane sugar is isotonic with a 1% solution of substance X.

### Step-by-Step Solution:

1. **Understanding the Concentration of Cane Sugar:**
- A 5% solution means there are 5 grams of cane sugar in 100 mL of solution.
- The molecular weight of cane sugar (sucrose) is given as 342 g/mol.

2. **Calculating the Molarity of Cane Sugar:**
- First, we need to find the number of moles of cane sugar in 5 grams.
\[ \text{Moles of cane sugar} = \frac{\text{mass}}{\text{molecular weight}} = \frac{5 \text{ g}}{342 \text{ g/mol}} \approx 0.0146 \text{ mol} \]

3. **Finding the Molality of Cane Sugar:**
- To find molality (m), we need to know the mass of the solvent. Assuming the density of the solution is similar to water, we can approximate the mass of the solvent.
- Total mass of the solution = 100 g (5 g sugar + 95 g solvent).
- Mass of the solvent (water) = 100 g - 5 g = 95 g.
- Convert grams of solvent to kilograms:
\[ \text{mass of solvent} = 95 \text{ g} = 0.095 \text{ kg} \]
- Now calculate molality:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.0146 \text{ mol}}{0.095 \text{ kg}} \approx 0.1547 \text{ mol/kg} \]

4. **Understanding the 1% Solution of Substance X:**
- A 1% solution means there is 1 gram of substance X in 100 mL of solution.
- Assuming the density of the solution is similar to water, the mass of the solvent is approximately 99 g (100 g - 1 g).

5. **Calculating the Molality of Substance X:**
- The mass of the solvent for substance X is 99 g, which is 0.099 kg.
- Let the molecular weight of substance X be \( M_X \).
- The number of moles of substance X in 1 g is:
\[ \text{Moles of X} = \frac{1 \text{ g}}{M_X} \]
- Now we can calculate the molality of substance X:
\[ m_X = \frac{1/M_X}{0.099} = \frac{1}{0.099 M_X} \]

6. **Setting the Molalities Equal:**
- Since the solutions are isotonic, their molalities are equal:
\[ 0.1547 = \frac{1}{0.099 M_X} \]

7. **Solving for \( M_X \):**
- Rearranging the equation gives:
\[ M_X = \frac{1}{0.1547 \times 0.099} \approx 65.6 \text{ g/mol} \]

### Final Answer:
The molecular weight of substance X is approximately **65.6 g/mol**.