At `25^(@)C` aqueous solution of glucose (molecular weight `= 180 g mol^(-1)`) is isotonic with a 2% aqueous solution containing an unknown solute. What is the molecular weight of the unknown solute

Options

A. 60
B. 80
C. 72
D. 63

Explanation

To solve the problem, we need to find the molecular weight of the unknown solute that makes a 2% solution isotonic with a solution of glucose. Here’s the step-by-step solution:

### Step 1: Understand the concept of isotonic solutions
Isotonic solutions have the same osmotic pressure. This means that the number of solute particles in both solutions is equal.

### Step 2: Calculate the number of moles of glucose
Given that the molecular weight of glucose is 180 g/mol, we can calculate the number of moles of glucose in a 5 g sample.

\[ \text{Number of moles of glucose} = \frac{\text{mass of glucose}}{\text{molar mass of glucose}} = \frac{5 \, \text{g}}{180 \, \text{g/mol}} = \frac{5}{180} \, \text{mol} \]

### Step 3: Determine the concentration of the glucose solution
Since we are considering isotonic solutions, the concentration of glucose must equal the concentration of the unknown solute. The concentration can be expressed in terms of moles per liter (mol/L). However, we need to relate it to the unknown solute.

### Step 4: Analyze the 2% solution of the unknown solute
A 2% solution means that there are 2 grams of the unknown solute in 100 mL of solution. To find the number of moles of the unknown solute, we will denote its molecular weight as \( M \).

\[ \text{Number of moles of unknown solute} = \frac{\text{mass of unknown solute}}{\text{molar mass of unknown solute}} = \frac{2 \, \text{g}}{M \, \text{g/mol}} \]

### Step 5: Set the moles of glucose equal to the moles of the unknown solute
Since the two solutions are isotonic, we can set the moles of glucose equal to the moles of the unknown solute:

\[ \frac{5}{180} = \frac{2}{M} \]

### Step 6: Solve for the molecular weight \( M \)
Cross-multiply to solve for \( M \):

\[ 5M = 2 \times 180 \]

\[ 5M = 360 \]

\[ M = \frac{360}{5} = 72 \, \text{g/mol} \]

### Conclusion
The molecular weight of the unknown solute is **72 g/mol**.

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