Question

Calculate the osmotic pressure of a solution prepared by dissolving 65.0 g of Na2SO4 in enough water to make 500 mL of solution at 20°C. (Assume no ion pairing – in other words, assume that the electrolyte completely dissociates into its constituent ions.)

Answer 1

Answer:

66.0 atm

Explanation:

We can calculate the osmotic pressure (π) using the following expression.

$\pi = i \times M \times R \times T$

where,

• i: van 't Hoff index

• M: molarity

• R: ideal gas constant

• T: absolute temperature

Step 1: Calculate i

Sodium sulfate completely dissociates according to the following equation.

Na₂SO₄ ⇒ 2 Na⁺ + SO₄²⁻

Since it produces 3 ions, i = 3.

Step 2: Calculate M

We can calculate the molarity of Na₂SO₄ using the following expression.

$M = \frac{mass\ of\ solute }{molar\ mass\ of\ solute\ \times liters\ of\ solution} = \frac{65.0g}{142.04g/mol \times 0.500L} =0.915M$

Step 3: Calculate T

We will use the following expression.

K = °C + 273.15

K = 20°C + 273.15 = 293 K

Step 4: Calculate π

$\pi = 3 \times 0.915M \times \frac{0.08206atm.L}{mol.K} \times 293K =66.0 atm$