A solution of substance containing `1.05` g per 100 mL was found to
be isotonic with `3%` glucose solution . The molecular mass
of the substance is :

Options

A. 31.5
B. 6.3
C. 630
D. 63

Explanation

To find the molecular mass of the substance that is isotonic with a 3% glucose solution, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Isotonic Solutions**:
- Two solutions are isotonic if they have the same osmotic pressure. Therefore, we can set the osmotic pressures of the two solutions equal to each other.

2. **Osmotic Pressure Formula**:
- The osmotic pressure (π) is given by the formula:
\[ \pi = C \cdot R \cdot T \]
- Where \(C\) is the concentration in moles per liter, \(R\) is the gas constant, and \(T\) is the temperature in Kelvin. Since \(R\) and \(T\) are constants for both solutions, we can simplify our equation to:
\[ C_1 = C_2 \]

3. **Calculating Concentration of the Substance**:
- The concentration \(C_1\) of the substance can be calculated using:
\[ C_1 = \frac{W}{M \cdot V} \]
- Where \(W\) is the weight of the substance, \(M\) is the molecular mass, and \(V\) is the volume in liters.
- Given \(W = 1.05 \, \text{g}\) and \(V = 100 \, \text{mL} = 0.1 \, \text{L}\), we can express \(C_1\) as:
\[ C_1 = \frac{1.05}{M \cdot 0.1} = \frac{10.5}{M} \, \text{mol/L} \]

4. **Calculating Concentration of the 3% Glucose Solution**:
- For a 3% glucose solution, the weight of glucose in 100 mL is 3 g. The molecular mass of glucose (C6H12O6) is 180 g/mol.
- The concentration \(C_2\) can be calculated as:
\[ C_2 = \frac{3 \, \text{g}}{180 \, \text{g/mol} \cdot 0.1 \, \text{L}} = \frac{3}{18} = \frac{1}{6} \, \text{mol/L} \]

5. **Setting the Concentrations Equal**:
- Since the solutions are isotonic, we set \(C_1 = C_2\):
\[ \frac{10.5}{M} = \frac{1}{6} \]

6. **Solving for Molecular Mass (M)**:
- Cross-multiplying gives:
\[ 10.5 \cdot 6 = M \]
\[ M = 63 \, \text{g/mol} \]

### Final Answer:
The molecular mass of the substance is **63 g/mol**.