Rank four solutions by osmotic pressure

Answer: D ($$0.08\ \mathrm{M}\ Al_{2}(SO_{4})_{3}$$)

Osmotic pressure is given by $$\Pi=iMRT$$, where $$i$$ is the van't Hoff factor. Because the solutions are at the same temperature, their values of $$iM$$ can be compared. The values are $$(1)$$$${}(0.28\ \mathrm{M})$$$${}=0.28\ \mathrm{M}$$ for $$C_{6}H_{12}O_{6}$$, $$(2)$$$${}(0.16\ \mathrm{M})$$$${}=0.32\ \mathrm{M}$$ for $$NaCl$$, and $$(3)$$$${}(0.12\ \mathrm{M})$$$${}=0.36\ \mathrm{M}$$ for $$CaCl_{2}$$. Each formula unit of $$Al_{2}(SO_{4})_{3}$$ produces five ions, so its value is $$(5)$$$${}(0.08\ \mathrm{M})$$$${}=0.40\ \mathrm{M}$$. Therefore, the $$Al_{2}(SO_{4})_{3}$$ solution has the greatest osmotic pressure.