A 5% (w/V ) solution of cane sugar (molecular mass = 342) is isotonic with 1% (w/V) solution of a subtance X. The molecular mass of X is :

Options

A. (a) 34.2
B. (b) 171.2
C. (c) 68.4
D. (d) 136.8

Answer

To solve the problem, we need to determine the molecular mass of substance X, given that a 5% (w/V) solution of cane sugar is isotonic with a 1% (w/V) solution of substance X. Here’s the step-by-step solution:

### Step 1: Understand the Definition of Isotonic Solutions
Isotonic solutions have the same osmotic pressure. Therefore, the osmotic pressure of the cane sugar solution will equal the osmotic pressure of the substance X solution.

### Step 2: Calculate the Molarity of the Cane Sugar Solution
A 5% (w/V) solution means there are 5 grams of cane sugar (sucrose) in 100 mL of solution.

1. **Convert grams to moles**:
\[ \text{Moles of cane sugar} = \frac{\text{mass}}{\text{molecular mass}} = \frac{5 \, \text{g}}{342 \, \text{g/mol}} \approx 0.0146 \, \text{mol} \]

2. **Calculate the volume in liters**:
\[ \text{Volume} = 100 \, \text{mL} = 0.1 \, \text{L} \]

3. **Calculate molarity (C)**:
\[ C = \frac{\text{moles}}{\text{volume in L}} = \frac{0.0146 \, \text{mol}}{0.1 \, \text{L}} = 0.146 \, \text{mol/L} \]

### Step 3: Calculate the Molarity of the Substance X Solution
A 1% (w/V) solution means there are 1 gram of substance X in 100 mL of solution.

1. **Convert grams to moles**:
\[ \text{Moles of substance X} = \frac{1 \, \text{g}}{\text{molecular mass of X (mmX)}} \]

2. **Calculate molarity (C)**:
\[ C = \frac{\text{moles}}{\text{volume in L}} = \frac{1 \, \text{g}/\text{mmX}}{0.1 \, \text{L}} = \frac{10}{\text{mmX}} \, \text{mol/L} \]

### Step 4: Set Up the Equation for Isotonic Solutions
Since the osmotic pressures are equal, we can set up the equation:
\[ \pi_{\text{cane sugar}} = \pi_{\text{X}} \]
Using the formula for osmotic pressure:
\[ I \cdot C_{\text{cane sugar}} \cdot R \cdot T = I \cdot C_{\text{X}} \cdot R \cdot T \]
Since both solutions are non-electrolytes, the van 't Hoff factor (I) is 1 for both.

Thus, we have:
\[ 0.146 = \frac{10}{\text{mmX}} \]

### Step 5: Solve for the Molecular Mass of X
Rearranging the equation gives:
\[ \text{mmX} = \frac{10}{0.146} \approx 68.49 \, \text{g/mol} \]

### Conclusion
The molecular mass of substance X is approximately **68.4 g/mol**.