50 mg of Na₂SO₄ is dissolved in 3 litres of water at 27°C. What would be the osmotic pressure for…

$\mathrm{Na_2SO_4}$ dissociates completely into 3 ions, so $i = 3$. Molar mass $= 142\text{ g mol}^{-1}$; moles $= \frac{0.050}{142} = 3.52 \times 10^{-4}$; volume $= 3\text{ L}$; $T = 300\text{ K}$. $\pi = \frac{i\,n\,R\,T}{V} = \frac{3 \times 3.52\times10^{-4} \times 0.0821 \times 300}{3} = 8.67 \times 10^{-3}\text{ atm}$. (The printed key's $1.23$ atm does not follow from its own working: 50 mg is $0.050$ g.)