Question

At $ 25{}^\circ C, $ at 5% aqueous solution of glucose (molecular weight $ =180\,g\,mo{{l}^{-1}} $ ) is isotonic with a 2% aqueous solution containing and unknown solute. What is the molecular weight of the unknown solute?

• A. 60
• B. 80
• C. 72
• D. 63

Answer 1

The Correct Option is C

Since the two solutions are isotonic, they must have same concentrations in moles/litre. For glucose solution, concentration
$ =5\text{ }g/100\text{ }c{{m}^{3}} $ (given)
$ =50g/L $
$ \therefore $ $ \frac{50}{180}=\frac{20}{M} $
or $ M=72 $
( $ \because $ Molar mass of glucose $ =180\text{ }g\text{ }mo{{l}^{-1}} $ )
For unknown substance, concentration
$ =2\text{ }g/100\text{ }c{{m}^{3}} $ (given)
$ =20g/L=\frac{20}{M}mol/L $
$ \therefore $ $ \frac{50}{180}=\frac{20}{M} $ or $ M=72 $

Answer 2

To find the molecular weight of the unknown solute, we can use the concept of isotonic solutions. Isotonic solutions have the same osmotic pressure, which implies that their concentrations in terms of molarity are equal when dissolved in water.

Given:
- Temperature: \( 25^\circ \text{C} \)
- Molecular weight of glucose: \( 180 \, \text{g/mol} \)
- 5% aqueous solution of glucose
- 2% aqueous solution of the unknown solute

Let's denote:
- Molecular weight of the unknown solute by \( M_u \)

A 5% solution means 5 grams of solute in 100 grams of solution. Since the density of water is approximately \( 1 \, \text{g/mL} \), we can approximate that 100 grams of solution is approximately 100 mL (or 0.1 L) of solution.

Step 1: Calculate the molarity of the glucose solution
For the glucose solution:
- Mass of glucose = 5 grams
- Volume of solution = 0.1 L

The number of moles of glucose in the solution:
\[ \text{Moles of glucose} = \frac{\text{Mass of glucose}}{\text{Molecular weight of glucose}} = \frac{5 \, \text{g}}{180 \, \text{g/mol}} \]

\[ \text{Moles of glucose} = \frac{5}{180} = 0.02778 \, \text{mol} \]

Molarity of the glucose solution:
\[ \text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in L}} = \frac{0.02778 \, \text{mol}}{0.1 \, \text{L}} \]

\[ \text{Molarity} = 0.2778 \, \text{M} \]

Step 2: Calculate the molarity of the unknown solute solution
For the unknown solute:
- Mass of the unknown solute = 2 grams
- Volume of solution = 0.1 L

Let \( M_u \) be the molecular weight of the unknown solute. The number of moles of the unknown solute in the solution is:
\[ \text{Moles of unknown solute} = \frac{\text{Mass of unknown solute}}{\text{Molecular weight of unknown solute}} = \frac{2 \, \text{g}}{M_u} \]

Molarity of the unknown solute solution:
\[ \text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in L}} = \frac{\frac{2 \, \text{g}}{M_u}}{0.1 \, \text{L}} \]

\[ \text{Molarity} = \frac{2}{0.1 \, M_u} \]

\[ \text{Molarity} = \frac{20}{M_u} \]

Step 3: Set the molarities equal (since the solutions are isotonic)
\[ 0.2778 \, \text{M} = \frac{20}{M_u} \]

Solving for \( M_u \):
\[ M_u = \frac{20}{0.2778} \]

\[ M_u = 72 \, \text{g/mol} \]

Therefore, the molecular weight of the unknown solute is \( 72 \, \text{g/mol} \)