Calculate the osmotic pressure associated with 50.0 g of an enzyme of molecular weight 98 g/mol dissolved in water to give 2600 ml of solution at 30.0 oc ?

π = 4,882 atm

Explanation:

To calculate the osmotic pressure (π), the Van´t Hoff equation must be used, which is:

π x V = n x R x T

Where:

• π: Osmotic pressure, which is the difference between the levels of the solution and the pure solvent through a semipermeable membrane, which allows the passage of the solvent but not the solute

• V: Volume of the solution, in liters unit

• n: Number of moles of solute

• R: Constant of ideal gases, equal to 0.08206 L.atm / mol.K

• T: Absolute temperature, in Kelvin degrees

With the data you provide you can calculate the osmotic pressure by clearing it from the equation, we would be equal to:

π = (n x R x T) / V

However, all data must first be converted to the corresponding units in order to replace the values in the equation.

Solution volume ⇒ go from mL to L:

1000 mL of solution 1 L

2600 mL of solution X = 2.6 L

Calculation: 2600 mL x 1 L / 1000 mL = 2.6 L

Temperature ⇒ Go from ° C to K

T (K) = t (° C) + 273.15 = 30.0 ° C + 273.15 = 303.15 K

Number of moles of solute ⇒ It can be calculated since we have the mass of the enzyme and its molecular mass:

98.0 g of enzyme 1 mol

50.0 g of enzyme X = 0.510 moles

Calculation: 50.0 g x 1 mol / 98.0 g = 0.510 moles

Now, you can replace the values in the Van´t Hoff equation and you will get the result:

 π = (n x R x T) / V

π = (0.510 mol x 0.08206 L.atm / mol.K x 303.15 K) / 2.6 L = 4.882 atm

Therefore, the osmotic pressure will be 4,882 atm

i believe the answer is a.