3% solution of glucose is isotonic with 1% solution of a non-volatile non-electrolyte substance. The molecular mass of the substance would be

Options

A. 180
B. 360
C. 420
D. 60

Answer

To solve the problem, we need to find the molecular mass of a non-volatile non-electrolyte substance that is isotonic with a 3% solution of glucose. Here’s a step-by-step solution:

### Step 1: Understand the given concentrations
- A 3% glucose solution means there are 3 grams of glucose in 100 mL of solution.
- A 1% solution of the non-volatile non-electrolyte means there is 1 gram of the non-electrolyte in 100 mL of solution.

### Step 2: Calculate the molarity of the glucose solution
Molarity (C) is defined as the number of moles of solute per liter of solution.
- First, we need to calculate the number of moles of glucose.
- The molecular formula of glucose is C₆H₁₂O₆. The molar mass of glucose can be calculated as follows:
- Carbon (C): 12 g/mol × 6 = 72 g/mol
- Hydrogen (H): 1 g/mol × 12 = 12 g/mol
- Oxygen (O): 16 g/mol × 6 = 96 g/mol
- Total molar mass of glucose = 72 + 12 + 96 = 180 g/mol

Now, calculate the number of moles of glucose in 3 grams:
\[ \text{Number of moles of glucose} = \frac{\text{mass}}{\text{molar mass}} = \frac{3 \text{ g}}{180 \text{ g/mol}} = \frac{1}{60} \text{ moles} \]

Next, convert the volume from mL to L:
\[ 100 \text{ mL} = 0.1 \text{ L} \]

Now we can calculate the molarity of the glucose solution:
\[ C_{\text{glucose}} = \frac{\text{Number of moles}}{\text{Volume in L}} = \frac{\frac{1}{60}}{0.1} = \frac{1}{6} \text{ M} \]

### Step 3: Calculate the molarity of the non-electrolyte solution
For the non-electrolyte solution:
- The mass is 1 gram in 100 mL (0.1 L).
- Let the molar mass of the non-electrolyte be \( M \).

The number of moles of the non-electrolyte is:
\[ \text{Number of moles of non-electrolyte} = \frac{1 \text{ g}}{M \text{ g/mol}} \]

The molarity of the non-electrolyte solution is:
\[ C_{\text{non-electrolyte}} = \frac{\frac{1}{M}}{0.1} = \frac{10}{M} \text{ M} \]

### Step 4: Set the osmotic pressures equal
Since the solutions are isotonic, their osmotic pressures are equal:
\[ C_{\text{glucose}} = C_{\text{non-electrolyte}} \]
\[ \frac{1}{6} = \frac{10}{M} \]

### Step 5: Solve for the molar mass \( M \)
Cross-multiply to solve for \( M \):
\[ M = 10 \times 6 = 60 \text{ g/mol} \]

### Conclusion
The molecular mass of the non-volatile non-electrolyte substance is **60 g/mol**.

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