The osmotic pressure of blood is 5.65 atm at `27^(@) C`.The number of mol of glucose to be used per litre for an intravenous injection that is to have the same osmotic pressure as blood is

Options

A. 0.14
B. 0.22
C. 0.32
D. 0.56

Explanation

To solve the problem of determining the number of moles of glucose required to match the osmotic pressure of blood, we can follow these steps:

### Step 1: Understand the formula for osmotic pressure
The osmotic pressure (\( \pi \)) is given by the formula:
\[ \pi = CRT \]
where:
- \( C \) is the concentration in moles per liter (mol/L),
- \( R \) is the ideal gas constant,
- \( T \) is the temperature in Kelvin.

### Step 2: Convert the temperature to Kelvin
The temperature is given as \( 27^\circ C \). To convert this to Kelvin:
\[ T(K) = 27 + 273 = 300 \, K \]

### Step 3: Identify the values
From the problem, we have:
- Osmotic pressure (\( \pi \)) = 5.65 atm
- Volume (\( V \)) = 1 L (since we are calculating per liter)
- Gas constant (\( R \)) = 0.0821 atm L K\(^{-1}\) mol\(^{-1}\)

### Step 4: Rearrange the formula to find concentration
We need to find the concentration \( C \) in terms of moles (\( n \)):
\[ C = \frac{n}{V} \]
Substituting this into the osmotic pressure formula gives:
\[ \pi = \frac{n}{V}RT \]
Rearranging for \( n \):
\[ n = \frac{\pi V}{RT} \]

### Step 5: Substitute the known values
Now we substitute the known values into the equation:
\[ n = \frac{5.65 \, \text{atm} \times 1 \, \text{L}}{0.0821 \, \text{atm L K}^{-1} \text{mol}^{-1} \times 300 \, K} \]

### Step 6: Calculate the number of moles
Calculating the denominator:
\[ 0.0821 \times 300 = 24.63 \, \text{atm L mol}^{-1} \]
Now substituting this back into the equation for \( n \):
\[ n = \frac{5.65}{24.63} \approx 0.229 \, \text{mol} \]

### Step 7: Conclusion
Thus, the number of moles of glucose required per liter for the intravenous injection to match the osmotic pressure of blood is approximately:
\[ n \approx 0.229 \, \text{mol} \]