2.5 g of protein containing only glycine `(C_2H_5NO_2)` is dissolved in water to make 500 mL of solution. The osmotic pressure of this solution at 300 K is found to be `5.03 xx 10^(-3)` bar. The total number of glycine units present in the protein is __________
(Given : R=0.083 L bar `K^(-1) mol^(-1)`)

Explanation

To solve the problem, we will follow these steps:

### Step 1: Use the formula for osmotic pressure
The osmotic pressure (\( \pi \)) is given by the formula:
\[ \pi = CRT \]
where:
- \( C \) = concentration of the solution in moles per liter (mol/L)
- \( R \) = universal gas constant = 0.083 L bar K\(^{-1}\) mol\(^{-1}\)
- \( T \) = temperature in Kelvin

### Step 2: Rearrange the formula to find the concentration
We can rearrange the formula to find \( C \):
\[ C = \frac{\pi}{RT} \]

### Step 3: Substitute the known values
Given:
- \( \pi = 5.03 \times 10^{-3} \) bar
- \( R = 0.083 \) L bar K\(^{-1}\) mol\(^{-1}\)
- \( T = 300 \) K

Substituting these values into the equation:
\[ C = \frac{5.03 \times 10^{-3}}{0.083 \times 300} \]

### Step 4: Calculate the concentration
Calculating the denominator:
\[ 0.083 \times 300 = 24.9 \]
Now substituting back:
\[ C = \frac{5.03 \times 10^{-3}}{24.9} \approx 2.02 \times 10^{-4} \text{ mol/L} \]

### Step 5: Calculate the number of moles in 500 mL
To find the number of moles of glycine in 500 mL (0.5 L):
\[ \text{Number of moles} = C \times \text{Volume} = 2.02 \times 10^{-4} \times 0.5 = 1.01 \times 10^{-4} \text{ moles} \]

### Step 6: Calculate the molar mass of glycine
The molar mass of glycine \( (C_2H_5NO_2) \) can be calculated as follows:
- Carbon (C): 2 × 12.01 g/mol = 24.02 g/mol
- Hydrogen (H): 5 × 1.008 g/mol = 5.04 g/mol
- Nitrogen (N): 1 × 14.01 g/mol = 14.01 g/mol
- Oxygen (O): 2 × 16.00 g/mol = 32.00 g/mol

Total molar mass of glycine:
\[ 24.02 + 5.04 + 14.01 + 32.00 = 75.07 \text{ g/mol} \]

### Step 7: Calculate the total mass of glycine in the protein
Using the number of moles calculated:
\[ \text{Mass of glycine} = \text{Number of moles} \times \text{Molar mass} = 1.01 \times 10^{-4} \times 75.07 \approx 0.00757 \text{ g} \]

### Step 8: Calculate the total mass of protein
Given that the total mass of the protein is 2.5 g, we can find the number of glycine units:
\[ \text{Total number of glycine units} = \frac{\text{Total mass of protein}}{\text{Mass of one glycine unit}} = \frac{2.5 \text{ g}}{0.00757 \text{ g}} \approx 330.25 \]

Since the number of units must be a whole number, we round it to:
\[ \text{Total number of glycine units} = 330 \]

### Final Answer
The total number of glycine units present in the protein is **330**.

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