The osmotic pressure of 0.1 M sodium chloride solution at `27^@C` is

Options

A. 4.0 atm
B. 2.46 atm
C. 4.92 atm
D. 1.23 atm

Answer

To calculate the osmotic pressure of a 0.1 M sodium chloride (NaCl) solution at 27°C, we will use the formula for osmotic pressure (π):

\[ \pi = iCRT \]

Where:
- \( \pi \) = osmotic pressure
- \( i \) = van 't Hoff factor (number of particles the solute dissociates into)
- \( C \) = molarity of the solution
- \( R \) = universal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature in Kelvin

### Step 1: Determine the van 't Hoff factor (i)
Sodium chloride (NaCl) dissociates into two ions in solution:
\[ NaCl \rightarrow Na^+ + Cl^- \]
Thus, \( i = 2 \) because one formula unit of NaCl produces two ions.

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

### Step 3: Substitute values into the osmotic pressure formula
Now we can substitute the values into the osmotic pressure formula:
- \( i = 2 \)
- \( C = 0.1 \, M \)
- \( R = 0.0821 \, L·atm/(K·mol) \)
- \( T = 300 \, K \)

\[ \pi = iCRT = 2 \times 0.1 \, M \times 0.0821 \, L·atm/(K·mol) \times 300 \, K \]

### Step 4: Calculate the osmotic pressure
Now, we can perform the calculation:
\[ \pi = 2 \times 0.1 \times 0.0821 \times 300 \]
\[ \pi = 2 \times 0.1 \times 24.63 \]
\[ \pi = 4.926 \, atm \]

### Final Answer
The osmotic pressure of the 0.1 M sodium chloride solution at 27°C is approximately **4.93 atm**.

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