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 of finding the molecular weight of substance X that is isotonic with a 5% solution of cane sugar, we will follow these steps:

### Step 1: Understand the concept of isotonic solutions
Isotonic solutions have the same osmotic pressure. For non-electrolytes, the osmotic pressure can be expressed in terms of molarity (C) and the molecular weight (M).

### Step 2: Calculate the molarity of the cane sugar solution
Given:
- A 5% solution of cane sugar (sucrose) with a molecular weight of 342 g/mol.

Assuming we have 100 mL of this solution:
- Mass of cane sugar = 5 g

To find the number of moles of cane sugar:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molecular weight}} = \frac{5 \, \text{g}}{342 \, \text{g/mol}} \approx 0.0146 \, \text{mol} \]

Since the volume of the solution is 0.1 L (100 mL):
\[ \text{Molarity (C1)} = \frac{\text{Number of moles}}{\text{Volume in L}} = \frac{0.0146 \, \text{mol}}{0.1 \, \text{L}} = 0.146 \, \text{mol/L} \]

### Step 3: Calculate the molarity of substance X
For a 1% solution of substance X:
- Mass of substance X = 1 g

Let the molecular weight of substance X be M g/mol. The number of moles of substance X is:
\[ \text{Number of moles} = \frac{1 \, \text{g}}{M \, \text{g/mol}} = \frac{1}{M} \, \text{mol} \]

Since the volume of the solution is also 0.1 L:
\[ \text{Molarity (C2)} = \frac{\text{Number of moles}}{\text{Volume in L}} = \frac{1/M}{0.1} = \frac{10}{M} \, \text{mol/L} \]

### Step 4: Set the molarities equal to each other
Since the solutions are isotonic:
\[ C1 = C2 \]
\[ 0.146 = \frac{10}{M} \]

### Step 5: Solve for M
Rearranging gives:
\[ M = \frac{10}{0.146} \approx 68.5 \, \text{g/mol} \]

### Conclusion
The molecular weight of substance X is approximately **68.5 g/mol**.

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