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 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 Isotonic Solutions**:
Isotonic solutions have the same osmotic pressure. This means that the number of solute particles in both solutions must be equal.

2. **Calculate the Molarity of the Cane Sugar Solution**:
- A 5% (w/w) solution of cane sugar means there are 5 grams of cane sugar in 100 grams of solution.
- The molecular weight of cane sugar (sucrose) is given as 342 g/mol.
- Number of moles of cane sugar = mass / molar mass = 5 g / 342 g/mol = 0.0146 moles.

3. **Calculate the Mass of Solvent**:
- The mass of the solvent (water) in the solution = total mass of solution - mass of solute = 100 g - 5 g = 95 g = 0.095 kg.

4. **Calculate the Molality of the Cane Sugar Solution**:
- Molality (m) = number of moles of solute / mass of solvent in kg = 0.0146 moles / 0.095 kg = 0.1547 mol/kg.

5. **Calculate the Molarity of Substance X**:
- Since the solutions are isotonic, the molality of the 1% solution of substance X must also be 0.1547 mol/kg.
- A 1% (w/w) solution of substance X means there are 1 gram of substance X in 100 grams of solution.
- The mass of the solvent (water) in this case = 100 g - 1 g = 99 g = 0.099 kg.

6. **Calculate the Number of Moles of Substance X**:
- Let the molecular weight of substance X be M grams/mol.
- Number of moles of substance X = mass / molar mass = 1 g / M g/mol = 0.01 / M moles.

7. **Set Up the Equation for Molality**:
- From the previous calculations, we know:
\[ \text{Molality of X} = \frac{0.01/M}{0.099} = 0.1547 \]
- Rearranging gives:
\[ 0.01/M = 0.1547 \times 0.099 \]
- Calculate the right side:
\[ 0.01/M = 0.0153 \]

8. **Solve for M**:
- Rearranging gives:
\[ M = 0.01 / 0.0153 \approx 653.59 \text{ g/mol} \]

Thus, the molecular weight of substance X is approximately **653.59 g/mol**.