The osmotic pressure of a solution prepared by dissolving 50 mg of \(H_2SO_4\) in 2 liters of water at \(25^o C\) will be

0.0187 atm

The correct answer is option 2. 0.0187 atm.

To calculate the osmotic pressure of the solution, we can use the following formula:

\(\pi = i \times \frac{n}{V}× R × T\)

where:

\(\pi \) is the osmotic pressure

\(i\) is the van't Hoff factor [\(H_2SO_4 \rightleftharpoons 2H^+ + SO_4^{2-},\, \ i = 3\)]

\(V\) is the volume which is \(2L\)

\(R\) is the Universal Gas Constant which is equal to \(0.0821\, \ L\, \ atm\, \ K^{-1}mol^{-1}\)

\(M\) is the mass of \(H_2SO_4\) which is \(98\, \ gmol^{-1}\)

\(T\) is the temperature which is \(25^oC = 298 K\)

Now,

\(\pi = i \times \frac{n}{V}× R × T\)

or, \(\pi = i \times \frac{w}{M} \times \frac{1}{V}× R × T\)

or, \(\pi = 3 \times \frac{0.050}{92} \times \frac{1}{2}× 0.0821 × 298\)

or, \(\pi = \frac{3.66987}{196}\)

or, \(\pi = 0.0187\, \ atm\)