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