The osomotic pressure `pi` depends on the molar concentration of the solution `(pi=CRT)`. If two solutions are of equal solute concentration and, hence, have the same omotic pressure, they are said to be isotonic. If two solutions are of unequal osmotic pressures, the more concentrated solution is said to be hypertonic and the more diluted solution is described as hypotonic.
Osmosis is the major mechanism for transporting water upward in the plants. Answer the following questions:
What would be the percent strength of solution of urea that would be isotonic with `4.5%` solution of glucose?

Options

A. `4.5%`
B. `13.5%`
C. `1.5%`
D. `9%`

Answer

To find the percent strength of a urea solution that would be isotonic with a 4.5% solution of glucose, we can follow these steps:

### Step 1: Determine the molarity of the glucose solution
- Given that we have a 4.5% glucose solution, this means there are 4.5 grams of glucose in 100 mL of solution.
- The molecular weight of glucose (C6H12O6) is approximately 180 g/mol.

**Calculation:**
\[ \text{Moles of glucose} = \frac{\text{mass}}{\text{molecular weight}} = \frac{4.5 \, \text{g}}{180 \, \text{g/mol}} = 0.025 \, \text{mol} \]

### Step 2: Calculate the molarity of the glucose solution
- Since the volume of the solution is 100 mL (or 0.1 L), we can calculate the molarity (C) using the formula:
\[ C = \frac{\text{moles}}{\text{volume in liters}} = \frac{0.025 \, \text{mol}}{0.1 \, \text{L}} = 0.25 \, \text{mol/L} \]

### Step 3: Set up the equation for the urea solution
- The osmotic pressure (π) is the same for both solutions since they are isotonic. Therefore, we can set the molarity of the urea solution equal to that of the glucose solution.
- The molecular weight of urea (NH2CONH2) is approximately 60 g/mol.

### Step 4: Calculate the mass of urea needed for isotonicity
- Let \( W \) be the mass of urea in grams in 100 mL of solution. The molarity of the urea solution can be expressed as:
\[ C_{\text{urea}} = \frac{W}{60 \, \text{g/mol} \times 0.1 \, \text{L}} = \frac{W}{6} \]
- Setting the molarity of urea equal to that of glucose:
\[ \frac{W}{6} = 0.25 \]

**Solving for \( W \):**
\[ W = 0.25 \times 6 = 1.5 \, \text{g} \]

### Step 5: Calculate the percent strength of the urea solution
- The percent strength (weight/volume percentage) is calculated as:
\[ \text{Percent strength} = \left( \frac{\text{mass of solute}}{\text{volume of solution}} \right) \times 100 = \left( \frac{1.5 \, \text{g}}{100 \, \text{mL}} \right) \times 100 = 1.5\% \]

### Final Answer:
The percent strength of the urea solution that would be isotonic with a 4.5% solution of glucose is **1.5%**.

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