Question

Determine the osmotic pressure of a solution prepared by dissolving 0.025 g of K₂SO₄ in 2 litres of water at 25°C, assuming that K₂SO₄ is completely dissociated. 

(R = 0.0821 Lit-atm K⁻¹ mol⁻¹, mol. wt. of K₂SO₄ = 174 g mol⁻¹)

Answer 1

Given: Mass of K₂SO₄ (w) = 0.025 g

Volume of solution (V) = 2 L

Temperature (T) = 25°C = 298 K

Molar mass of K₂SO₄ = 174 g/mol

Gas constant (R) = 0.0821 L atm mol⁻¹ K⁻¹

Assuming complete dissociation of K₂SO₄

\[\ce{K2SO4 -> 2K+ + SO^2-_4}\]

⇒ i = 3

Moles of K₂SO₄ (n) = `0.025/174`

= 1.4368 × 10⁻⁴ mol

Concentration (C) = `n/V`

= `(1.4368 xx 10^-4)/2`

= 7.148 × 10⁻⁵ mol/L

By using Van’t Hoff equation for osmotic pressure

π = i CRT

= 3 × 7.148 × 10⁻⁵ × 0.0821 × 298

= 2.1552 × 10⁻⁴ × 0.0821 × 298

= 0.00528 atm

Answer 2

Given: Mass of K₂SO₄ (w) = 0.025 g

Volume of solution (V) = 2 L

Temperature (T) = 25°C = 298 K

Molar mass of K₂SO₄ = 174 g/mol

Gas constant (R) = 0.0821 L atm mol⁻¹ K⁻¹

When K₂SO₄ is dissolved in water, K⁺ and \[\ce{SO^{2-}_{4}}\] ions are produced.

\[\ce{K2SO4 -> 2K+ + SO^2-_4}\]

Total number of ions produced = 3

∴ Van’t Hoff factor (i) = 3

Applying the following relation for osmotic pressure (π),

π = `i n/v RT`

= `i xx w/M xx 1/v xx RT`

= `3 xx 0.025/174 xx 1/2 xx 0.0821 xx 298`

= 5.27 × 10⁻³ atm

Hence, the osmotic pressure of a solution will be 5.27 × 10⁻³ atm.