Question

Determine the osmotic pressure of a solution prepared by dissolving 25 mg of \(K_2SO_4\) in 2 litre of water at \(25\degree C\), assuming that it is completely dissociated.

Answer 1

The correct answer is: \(5.27 \times 10^{-3}atm\)
When \(K_2SO_4\) is dissolved in water \(K^+\) and \(SO_4^{2-}\) ions are produced.
\(K_2SO_4→2K^++SO_4^{2-}\)
Total number of ions produced = 3
\(\therefore i =3 \)
Given, 
\(w = 25 mg = 0.025 g \)
\(V = 2 L \)
\(T =\) \(25 \degree C\) \(= (25 + 273) K = 298 K \)
Also, we know that: 
\(R = 0.0821\, L\, atm\, K^{-1}mol^{-1 }\)
\(M = (2 × 39) + (1 × 32) + (4 × 16) = 174 g mol^{-1 }\)
Appling the following relation,
\(π=i\frac{n}{v}RT\)
\(=i\frac{w}{M}\frac{1}{v}RT\)
\(=3 \times \frac{0.025}{174} \times \frac{1}{2} \times 0.0821 \times 298\)
\(=5.27 \times 10^{-3}atm\)