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

0.527 atm

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

\(\pi = \text{van’t Hoff factor} × \text{ molar concentration }× R × T\)

where:

\(\pi \) is the osmotic pressure in atmospheres (atm) van’t Hoff factor \((i)\) is a factor that accounts for the dissociation of the solute. For sulfuric acid, \(i = 2\), because it dissociates into two ions: \(H^+\) and \(SO_4^{2−}\). molar concentration \((M)\) is the moles of solute per liter of solution. \(R\) is the gas constant, which is \(0.0821 L·atm/mol·K\). \(T\) is the temperature in Kelvin.

First, we need to calculate the molar concentration of the solution. To do this, we need to know the number of moles of solute in 2 liters of solution. We can use the following formula:

\(n(solute) = \frac{m(solute)}{M(solute)}\)

where:

n(solute) is the moles of solute m(solute) is the mass of solute, which is 50 mg in this case. M(solute) is the molar mass of solute, which is \(98.08 g/mol\) for \(H_2SO_4\).

Plugging in the values, we get:

\(n(solute) = 50 mg × (1 g/1000 mg) / 98.08 g/mol\)

\(n(solute) = 5.10 × 10^{−4} mol/L\)

Next, we need to calculate the temperature in Kelvin. We are given that the temperature is \(25^oC\), which is equal to 298 K.

Now we can plug all of the values into the formula for osmotic pressure:

\(\pi = 2 × 5.10 × 10^{−4} mol/L × 0.0821 L·atm/mol·K × 298 K\)

\(\pi = 0.527 atm\)

Therefore, the osmotic pressure of the solution is 0.527 atm. The correct answer is 4. 0.527 atm.