20 g of a substance were dissolved in 500 mL of water and the osmotic pressure of the solution was found to be 600 mm of mercury at 15°C. The molecular weight of a substance is:

1198

The correct answer is option 4. 1198.

To determine the molecular weight of the substance, we can use the formula for osmotic pressure:

\(\pi = \frac{nRT}{V}\)

where,

\(\pi\) is the osmotic pressure (600 mmHg in this case, which we need to convert to atm).

\(n\) is the number of moles of the solute.

\(R\) is the gas constant (\(0.0821 \, \text{L} \, \text{atm} \, \text{mol}^{-1} \, \text{K}^{-1}\)).

\(T\) is the temperature in Kelvin.

\(V\) is the volume of the solution in liters.

We know,

\(1\, \ atm = 760\, \ mmHg\)

\(\pi = \frac{600 \, \text{mmHg}}{760 \, \text{mmHg/atm}} \approx 0.789 \, \text{atm}\)

Given,

\(T = 15^\circ C + 273 = 288 \, \text{K}\)

We know,

\(\pi = \frac{nRT}{V}\)

or, \(n = \frac{\pi V}{RT}\)

Given:

\(V = 500 \, \text{mL} = 0.5 \, \text{L}\)

\(R = 0.0821 \, \text{L} \, \text{atm} \, \text{mol}^{-1} \, \text{K}^{-1}\)

Substitute the values:

\(n = \frac{0.789 \, \text{atm} \times 0.5 \, \text{L}}{0.0821 \, \text{L} \, \text{atm} \, \text{mol}^{-1} \, \text{K}^{-1} \times 288 \, \text{K}}\)

\(n \approx \frac{0.3945 \, \text{atm} \times \text{L}}{23.6448 \, \text{atm} \, \text{L} \, \text{mol}^{-1}}\)

\(n \approx 0.01668 \, \text{mol}\)

Thus, the molecular weight (\(M\)) is

\(M = \frac{\text{mass of substance}}{n} = \frac{20 \, \text{g}}{0.01668 \, \text{mol}} \approx 1198 \, \text{g/mol}\)

Conclusion: The molecular weight of the substance is 1198 g/mol.