Question

What is the osmotic pressure of a solution formed by dissolving 44.5 mg of aspirin (C9H8O4) in 0.355 L of water at 25 ∘C ? Activate to select the appropriates template from the following choices. Operate up and down arrow for selection and press enter to choose the input value typeActivate to select the appropriates symbol from the following choices. Operate up and down arrow for selection and press enter to choose the input value type π = atm

Answer 1

The osmotic pressure of the solution formed by dissolving 44.5 mg of aspirin in 0.355 L of water at 25 ∘C is approximately 5.72 atm

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

π = iMRT

Where:

π = osmotic pressure

i = van't Hoff factor (for aspirin, i = 1 because it does not dissociate in water)

M = molarity of the solution

R = ideal gas constant (0.0821 L.atm/mol.K)

T = temperature in Kelvin

First, we need to calculate the molarity of the aspirin solution:

Molarity (M) = moles of solute / liters of solution

First, we need to convert the mass of aspirin to moles:

44.5 mg * (1 g / 1000 mg) * (1 mol / 180.16 g) = 0.000247 mol

Now we can calculate the molarity:

Molarity (M) = 0.000247 mol / 0.355 L = 0.000695 M

Now we can calculate the osmotic pressure using the formula:

π = (1)(0.000695 M)(0.0821 L.atm/mol.K)(25 + 273 K)

≈ 5.72 atm

So, the osmotic pressure of the solution formed by dissolving 44.5 mg of aspirin in 0.355 L of water at 25 ∘C is approximately 5.72 atm.

The probable question may be:

What is the osmotic pressure of a solution formed by dissolving 44.5 mg of aspirin (C9H8O4) in 0.355 L of water at 25 ∘C ?