The osmotic pressure of blood is 8.21 atm at `37^(@)C`. How much glucose would be used for an injection that is at the same osmotic pressure as blood?

Explanation

To find out how much glucose is needed for an injection that has the same osmotic pressure as blood, we can use the formula for osmotic pressure:

\[ \Pi = iCRT \]

Where:
- \(\Pi\) = osmotic pressure (in atm)
- \(i\) = van 't Hoff factor (for glucose, \(i = 1\) since it does not dissociate in solution)
- \(C\) = molarity of the solution (in mol/L)
- \(R\) = ideal gas constant (0.0821 L·atm/(K·mol))
- \(T\) = temperature in Kelvin

### Step 1: Convert the temperature from Celsius to Kelvin
The temperature in Kelvin can be calculated using the formula:

\[ T(K) = T(°C) + 273.15 \]

For \(37°C\):

\[ T = 37 + 273.15 = 310.15 \, K \]

### Step 2: Rearrange the osmotic pressure formula to find molarity (C)
We can rearrange the formula to solve for \(C\):

\[ C = \frac{\Pi}{iRT} \]

Substituting the known values:

\[ C = \frac{8.21 \, \text{atm}}{(1)(0.0821 \, \text{L·atm/(K·mol)})(310.15 \, K)} \]

### Step 3: Calculate the molarity (C)
Now, we can calculate \(C\):

\[ C = \frac{8.21}{0.0821 \times 310.15} \]

Calculating the denominator:

\[ 0.0821 \times 310.15 \approx 25.485 \]

Now substituting back:

\[ C \approx \frac{8.21}{25.485} \approx 0.322 \, \text{mol/L} \]

### Step 4: Calculate the amount of glucose needed
To find the mass of glucose required, we can use the formula:

\[ \text{mass} = C \times V \times M \]

Where:
- \(V\) = volume of the solution (assume 1 L for simplicity)
- \(M\) = molar mass of glucose (C₆H₁₂O₆) = 180.18 g/mol

Assuming we are preparing 1 L of solution:

\[ \text{mass} = 0.322 \, \text{mol/L} \times 1 \, \text{L} \times 180.18 \, \text{g/mol} \]

Calculating the mass:

\[ \text{mass} \approx 0.322 \times 180.18 \approx 58.1 \, \text{g} \]

### Final Answer
The amount of glucose needed for an injection that has the same osmotic pressure as blood is approximately **58.1 grams**.

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