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 the solution in liters
- \(R\) = universal 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 \(\pi = 500 \, \text{cm Hg}\).

To convert this to atm:
\[ \pi = \frac{500 \, \text{cm Hg}}{76 \, \text{cm Hg/atm}} = \frac{500}{76} \, \text{atm} \approx 6.58 \, \text{atm} \]

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

Convert this to liters:
\[ V = 100 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} = 0.1 \, \text{L} \]

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

Convert this to Kelvin:
\[ T = 27 + 273.15 = 300.15 \, \text{K} \approx 300 \, \text{K} \]

### Step 4: Calculate the number of moles of solute
We know that:
\[ \pi V = nRT \]
Rearranging gives:
\[ n = \frac{\pi V}{RT} \]

Substituting the values:
\[ n = \frac{(6.58 \, \text{atm})(0.1 \, \text{L})}{(0.0821 \, \text{L·atm/(K·mol)})(300 \, \text{K})} \]

Calculating:
\[ n = \frac{0.658}{24.63} \approx 0.0267 \, \text{mol} \]

### Step 5: Calculate the molecular weight of the solute
Molecular weight \(M\) can be calculated using:
\[ M = \frac{m}{n} \]
Where \(m\) is the mass of the solute (4 g).

Substituting the values:
\[ M = \frac{4 \, \text{g}}{0.0267 \, \text{mol}} \approx 149.06 \, \text{g/mol} \]

### Final Answer
The molecular weight of the solute is approximately \(149.06 \, \text{g/mol}\).

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