Calculate the osmotic pressure of this solution.
69.7 grams of a solute with a molecular mass of 2790 grams are
dissolved in enough water to make 1.00 dm' of solution at 20 °c.
what is the osmotic pressure of the solution?

The osmotic pressure of the solution is 0,60 atm

Explanation:

This is the formula for osmotic pressure.

π = n . R . T . i

where n is the molar concentration of solute (mol/L), R is the ideal gas constant, and T is the temperature in Kelvins. The i means  the degree of dissociation of the solute, in this case it does not clarify it so we assume that it is a non-volatile solute, therefore the i is worth 1 .

1 dm³ = 1 L

So we have to get the moles.

Mass/ Molecular mass = Moles

69,7 g / 2790 g/m = 0,025 moles

T° in K = T° in °C + 273 ---> 20 + 273 = 293

π = 0,025 moles/L .  0,082 L.atm/K.mol . 293K . 1

π = 0,60 atm

the answer to your question is c