The osmotic pressure of 0.0400 mol L-1 NaNO3 at 25.0oC is:Question 8Select one:100 kPa350 kPa198 kPa99 kPa18.5 kPa

To solve this problem, we need to use the formula for osmotic pressure:

Π = nRT/V

where: Π is the osmotic pressure, n is the number of moles of solute, R is the ideal gas constant, T is the temperature in Kelvin, V is the volume in liters.

Given: n = 0.0400 mol R = 0.0821 L·atm/(K·mol) (This is the value of R in these units) T = 25.0°C = 298.15 K (Converted from Celsius to Kelvin) V = 1 L (Since the concentration is given in mol/L, we can assume the volume is 1 L)

Substituting these values into the formula gives:

Π = (0.0400 mol)(0.0821 L·atm/(K·mol))(298.15 K) / 1 L

Π = 0.976 atm

However, the options are given in kPa. We know that 1 atm = 101.325 kPa, so we convert:

Π = 0.976 atm * 101.325 kPa/atm = 98.89 kPa

So, the closest answer would be 99 kPa.