The osmotic pressure of blood is 7.65 atm, at 310K. An aqueous solution of glucose that will be isotonic with blood is of what wt./vol?

To determine the weight/volume percentage of a glucose solution that is isotonic with blood, we can use the concept of osmotic pressure. The osmotic pressure (\( \Pi \)) of a solution can be calculated using the formula:

Osmotic Pressure Formula

\( \Pi = iCRT \)

• i = van 't Hoff factor (for glucose, \( i = 1 \))
• C = molarity of the solution (in mol/L)
• R = ideal gas constant (0.0821 L·atm/(K·mol))
• T = temperature in Kelvin (310 K)

Calculating Molarity

Given that the osmotic pressure of blood is 7.65 atm, we can rearrange the formula to find the molarity (C):

\( C = \frac{\Pi}{iRT} \)

Substituting the values:

\( C = \frac{7.65 \, \text{atm}}{(1)(0.0821 \, \text{L·atm/(K·mol)})(310 \, \text{K})} \)

Calculating this gives:

\( C \approx 0.299 \, \text{mol/L} \)

Finding Weight/Volume Percentage

Next, we need to convert molarity to weight/volume percentage. The molar mass of glucose (C₆H₁₂O₆) is approximately 180 g/mol. To find the weight of glucose in 1 liter of solution:

Weight of glucose = Molarity × Molar mass = 0.299 mol/L × 180 g/mol = 53.82 g

To find the weight/volume percentage:

Weight/Volume % = \( \frac{\text{Weight of solute (g)}}{\text{Volume of solution (mL)}} \times 100 \)

Thus, for 53.82 g in 1000 mL:

Weight/Volume % = \( \frac{53.82 \, \text{g}}{1000 \, \text{mL}} \times 100 \approx 5.38\% \)

Final Result

The weight/volume percentage of the glucose solution that will be isotonic with blood is approximately 5.38%.