Osmotic pressure of blood is 7.40 atm, at `27^(@)`C. Number of moles of glucose to be used per litre for an intravenous injection that is to have same osmotic pressure of blood is :

Options

A. (a) 0.3
B. (b) 0.2
C. (c) 0.1
D. (d) 0.4

Answer

To find the number of moles of glucose needed per liter for an intravenous injection that matches the osmotic pressure of blood, we can follow these steps:

### Step 1: Understand the Osmotic Pressure Formula
The osmotic pressure (π) is given by the formula:
\[ \pi = iCRT \]
where:
- \( \pi \) = osmotic pressure
- \( i \) = Van't Hoff factor (number of particles the solute dissociates into)
- \( C \) = concentration in moles per liter (mol/L)
- \( R \) = universal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature in Kelvin

### Step 2: Identify Given Values
From the question, we have:
- Osmotic pressure of blood, \( \pi = 7.40 \, \text{atm} \)
- Temperature, \( T = 27^\circ C \)

### Step 3: Convert Temperature to Kelvin
To convert Celsius to Kelvin, use the formula:
\[ T(K) = T(°C) + 273 \]
So,
\[ T = 27 + 273 = 300 \, K \]

### Step 4: Determine the Van't Hoff Factor for Glucose
Glucose (C₆H₁₂O₆) is a non-electrolyte, meaning it does not dissociate in solution. Therefore, the Van't Hoff factor \( i = 1 \).

### Step 5: Substitute Values into the Osmotic Pressure Formula
Now we can substitute the known values into the osmotic pressure formula:
\[ 7.40 = (1) \cdot C \cdot (0.0821) \cdot (300) \]

### Step 6: Solve for Concentration \( C \)
Rearranging the equation to solve for \( C \):
\[ C = \frac{7.40}{0.0821 \cdot 300} \]

Calculating the denominator:
\[ 0.0821 \cdot 300 = 24.63 \]
Now substituting back to find \( C \):
\[ C = \frac{7.40}{24.63} \approx 0.300 \, \text{mol/L} \]

### Step 7: Conclusion
The number of moles of glucose needed per liter for the intravenous injection is approximately:
\[ \text{Number of moles} = 0.300 \, \text{mol} \]

Thus, the answer is **0.300 moles of glucose per liter**.

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