A solution containing `4 g` of a non-volatile organic solute per `100 mL` was found to have an osmotic pressure equal to `500 cm` of mercury at `27^(@)C`. The molecular weight of solute is

Options

A. 14.97
B. 149.7
C. 1697
D. 1.497

Explanation

To find the molecular weight of the non-volatile organic solute, we can use the formula for osmotic pressure:

\[ \pi = \frac{n}{V}RT \]

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

### Step 1: Convert osmotic pressure from cm of mercury to atm
Given osmotic pressure \(\pi = 500 \, \text{cm Hg}\):

\[ \pi = \frac{500}{76} \, \text{atm} \]

### Step 2: Convert volume from mL to L
Given volume of solution \(V = 100 \, \text{mL}\):

\[ V = \frac{100}{1000} \, \text{L} = 0.1 \, \text{L} \]

### Step 3: Convert temperature from Celsius to Kelvin
Given temperature \(T = 27^\circ C\):

\[ T = 27 + 273 = 300 \, \text{K} \]

### Step 4: Calculate the number of moles of solute (n)
We know that:

\[ n = \frac{mass}{molecular \, weight} = \frac{W_B}{M_B} \]

Where:
- \(W_B\) = mass of solute = 4 g
- \(M_B\) = molecular weight of solute (unknown)

### Step 5: Substitute values into the osmotic pressure formula
Substituting the known values into the osmotic pressure equation:

\[ \frac{500}{76} = \frac{W_B}{M_B \cdot V} \cdot R \cdot T \]

Substituting the values we have:

\[ \frac{500}{76} = \frac{4}{M_B \cdot 0.1} \cdot 0.0821 \cdot 300 \]

### Step 6: Rearranging the equation to solve for \(M_B\)
Rearranging gives us:

\[ M_B = \frac{4 \cdot 0.0821 \cdot 300}{\frac{500}{76} \cdot 0.1} \]

### Step 7: Calculate the molecular weight
Now, let's calculate \(M_B\):

1. Calculate the left side:
\[ \frac{500}{76} \approx 6.5789 \, \text{atm} \]

2. Substitute into the equation:
\[ M_B = \frac{4 \cdot 0.0821 \cdot 300}{6.5789 \cdot 0.1} \]

3. Calculate:
\[ M_B = \frac{4 \cdot 24.63}{0.65789} \approx \frac{98.52}{0.65789} \approx 149.7 \, \text{g/mol} \]

### Final Answer
The molecular weight of the solute is approximately **149.7 g/mol**.

---