Sea water is `3.5%`by mass of a salt and has a density `1.04gcm^(-3)` at `293K`.Assuming the salt to be sodium chloride ,calculate the osmotic pressure of sea water.Assume complete ionisation of the salt-

Options

A. 25.45 atm
B. 11.56 atm
C. 29.98 atm
D. 30.20 atm

Explanation

To calculate the osmotic pressure of seawater, we will follow these steps:

### Step 1: Identify the given data
- Percentage by mass of NaCl in seawater = 3.5%
- Density of seawater = 1.04 g/cm³
- Temperature (T) = 293 K
- Molar mass of NaCl = 58.5 g/mol

### Step 2: Calculate the mass of NaCl in 100 g of seawater
Since seawater is 3.5% NaCl by mass, in 100 g of seawater, the mass of NaCl is:
\[ \text{Mass of NaCl} = 3.5 \, \text{g} \]

### Step 3: Calculate the volume of seawater
Using the density formula:
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \implies \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]
Substituting the values:
\[ \text{Volume} = \frac{100 \, \text{g}}{1.04 \, \text{g/cm}^3} = 96.15 \, \text{cm}^3 = 0.09615 \, \text{L} \]

### Step 4: Calculate the number of moles of NaCl
Using the formula:
\[ \text{Number of moles} = \frac{\text{Mass}}{\text{Molar mass}} \]
Substituting the values:
\[ \text{Number of moles of NaCl} = \frac{3.5 \, \text{g}}{58.5 \, \text{g/mol}} = 0.0599 \, \text{mol} \]

### Step 5: Calculate the molarity (C) of the NaCl solution
Molarity is given by:
\[ C = \frac{\text{Number of moles}}{\text{Volume in L}} \]
Substituting the values:
\[ C = \frac{0.0599 \, \text{mol}}{0.09615 \, \text{L}} = 0.623 \, \text{mol/L} \]

### Step 6: Calculate the van 't Hoff factor (i)
For NaCl, which dissociates into Na⁺ and Cl⁻:
\[ i = 2 \]

### Step 7: Use the osmotic pressure formula
The osmotic pressure (π) is given by:
\[ \pi = iCRT \]
Where:
- \( R = 0.0821 \, \text{L·atm/(K·mol)} \)
- \( T = 293 \, \text{K} \)

Substituting the values:
\[ \pi = 2 \times 0.623 \, \text{mol/L} \times 0.0821 \, \text{L·atm/(K·mol)} \times 293 \, \text{K} \]

### Step 8: Calculate the osmotic pressure
Calculating the above expression:
\[ \pi = 2 \times 0.623 \times 0.0821 \times 293 = 29.93 \, \text{atm} \]

### Final Answer
The osmotic pressure of seawater is approximately:
\[ \pi \approx 29.98 \, \text{atm} \]