An aqueous solution containing `21.6 mg` at a solute in `100 ml` of solution has an osmotic pressure of `3.70 mm` of `Hg` at `25^(@)C`. The molecular wt of solute in `g//mol` is

Options

A. 1085
B. 9035
C. 1355
D. 700

Explanation

To find the molecular weight of the solute in the given aqueous solution, we can use the formula for osmotic pressure:

\[ \Pi V = nRT \]

Where:
- \(\Pi\) = osmotic pressure
- \(V\) = volume of the solution in liters
- \(n\) = number of moles of solute
- \(R\) = ideal gas constant (0.0821 L·atm/(K·mol))
- \(T\) = temperature in Kelvin

### Step 1: Convert the given values to appropriate units
- Osmotic pressure (\(\Pi\)) is given as \(3.70 \, \text{mm Hg}\). We need to convert this to atm:
\[ \Pi = \frac{3.70 \, \text{mm Hg}}{760 \, \text{mm Hg/atm}} = 0.004868 \, \text{atm} \]
- Volume (\(V\)) is given as \(100 \, \text{ml}\). We need to convert this to liters:
\[ V = \frac{100 \, \text{ml}}{1000} = 0.1 \, \text{L} \]
- Temperature (\(T\)) is given as \(25^\circ C\). We need to convert this to Kelvin:
\[ T = 25 + 273 = 298 \, \text{K} \]

### Step 2: Calculate the number of moles (\(n\))
We know that the number of moles \(n\) can be expressed as:
\[ n = \frac{W}{M} \]
Where:
- \(W\) = mass of solute in grams
- \(M\) = molecular weight of the solute in g/mol

The mass of solute is given as \(21.6 \, \text{mg}\), which we convert to grams:
\[ W = 21.6 \, \text{mg} = 21.6 \times 10^{-3} \, \text{g} \]

### Step 3: Substitute values into the osmotic pressure equation
Now substituting the values into the osmotic pressure equation:
\[ \Pi V = \frac{W}{M} RT \]
Substituting the known values:
\[ 0.004868 \, \text{atm} \times 0.1 \, \text{L} = \frac{21.6 \times 10^{-3} \, \text{g}}{M} \times 0.0821 \, \text{L·atm/(K·mol)} \times 298 \, \text{K} \]

### Step 4: Solve for \(M\)
Rearranging the equation to solve for \(M\):
\[ M = \frac{21.6 \times 10^{-3} \times 0.0821 \times 298}{0.004868 \times 0.1} \]

Calculating the right-hand side:
1. Calculate the numerator:
\[ 21.6 \times 10^{-3} \times 0.0821 \times 298 \approx 0.529 \]
2. Calculate the denominator:
\[ 0.004868 \times 0.1 \approx 0.000487 \]
3. Now divide:
\[ M \approx \frac{0.529}{0.000487} \approx 1085.2 \, \text{g/mol} \]

### Final Answer
The molecular weight of the solute is approximately \(1085 \, \text{g/mol}\).

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