The osmotic pressure of blood is `7.65`atm at `310K`. An aqueous solution of glucose which is isotonic with blood has the percentage (wt./volume)

Options

A. `5.41%`
B. `3.54%`
C. `4.53%`
D. `53.4%`

Answer

To solve the problem, we need to find the percentage (weight/volume) of an aqueous glucose solution that is isotonic with blood, given that the osmotic pressure of blood is 7.65 atm at a temperature of 310 K.

### Step-by-Step Solution:

1. **Understanding Osmotic Pressure**:
The osmotic pressure (π) of a solution can be calculated using the formula:
\[ \pi = cRT \]
where:
- \( \pi \) = osmotic pressure (in atm)
- \( c \) = molarity of the solution (in moles per liter)
- \( R \) = universal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature in Kelvin

2. **Setting Up the Equation**:
Since the osmotic pressure of the glucose solution is equal to that of blood, we can set up the equation:
\[ 7.65 = c \times 0.0821 \times 310 \]

3. **Solving for Molarity (c)**:
Rearranging the equation to solve for \( c \):
\[ c = \frac{7.65}{0.0821 \times 310} \]
Now, calculate the value:
\[ c = \frac{7.65}{25.451} \approx 0.301 \text{ moles per liter} \]

4. **Finding the Mass of Glucose**:
To find the mass of glucose in grams, we use the molarity and the molecular weight of glucose (C6H12O6), which is 180 g/mol:
\[ \text{Mass} = c \times \text{Molar Mass} \times \text{Volume} \]
Assuming we are calculating for 1 liter of solution:
\[ \text{Mass} = 0.301 \, \text{mol/L} \times 180 \, \text{g/mol} \times 1 \, \text{L} \approx 54.18 \, \text{g} \]

5. **Calculating Percentage (wt/volume)**:
The percentage weight/volume (w/v) is calculated as:
\[ \text{Percentage (w/v)} = \left( \frac{\text{mass of solute (g)}}{\text{volume of solution (mL)}} \right) \times 100 \]
Since we have 54.18 g of glucose in 1000 mL (1 L):
\[ \text{Percentage (w/v)} = \left( \frac{54.18}{1000} \right) \times 100 \approx 5.418 \% \]

6. **Final Answer**:
Therefore, the percentage (weight/volume) of the glucose solution that is isotonic with blood is approximately **5.41%**.