10%(w/v) solution of glucose is isotonic with 4% (w/v) solution of
non - volatile solution then molar mass of non - volatile solute
will be :-

Options

A. 36
B. 72
C. 54
D. 63

Explanation

To solve the problem, we need to determine the molar mass of a non-volatile solute that is isotonic with a 10% (w/v) glucose solution. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Concept of Isotonic Solutions
Isotonic solutions have the same osmotic pressure, which means they contain the same number of solute particles per unit volume. For two solutions to be isotonic, the number of moles of solute in each solution must be equal.

### Step 2: Determine the Molar Concentration of Glucose
Given that we have a 10% (w/v) solution of glucose, this means there are 10 grams of glucose in 100 mL of solution.

- **Molar mass of glucose (C6H12O6)** = 180 g/mol
- **Number of moles of glucose (n_glucose)** = mass (g) / molar mass (g/mol)
\[ n_{\text{glucose}} = \frac{10 \text{ g}}{180 \text{ g/mol}} = \frac{1}{18} \text{ mol} \]

### Step 3: Determine the Molar Concentration of the Non-volatile Solute
Next, we have a 4% (w/v) solution of a non-volatile solute. This means there are 4 grams of the solute in 100 mL of solution.

- Let the molar mass of the non-volatile solute be \( x \) g/mol.
- **Number of moles of the non-volatile solute (n_solute)** = mass (g) / molar mass (g/mol)
\[ n_{\text{solute}} = \frac{4 \text{ g}}{x \text{ g/mol}} \]

### Step 4: Set Up the Equation for Isotonic Solutions
Since the two solutions are isotonic, we can set the number of moles equal to each other:
\[ n_{\text{glucose}} = n_{\text{solute}} \]
\[ \frac{10}{180} = \frac{4}{x} \]

### Step 5: Solve for \( x \)
Cross-multiply to solve for \( x \):
\[ 10x = 4 \times 180 \]
\[ 10x = 720 \]
\[ x = \frac{720}{10} = 72 \text{ g/mol} \]

### Conclusion
The molar mass of the non-volatile solute is **72 g/mol**.

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