Answer
$34.2atm$
Work Step by Step
We can calculate the required osmotic pressure as follows:
$M=(\frac{28.5g\space Mg_3(C_6H_5O_7)_2}{235.0\times 10^{-3}Lsoln})(\frac{1molMg_3(C_6H_5O_7)_2}{451.114g\space Mg_3(C_6H_5O_7)_2})=0.26884M$
Now $\pi=MRTi$
We plug in the known values to obtain:
$\pi=(\frac{0.26884mol}{L})(\frac{0.08206L.atm}{mol.K})(310.15K)(5)$
$\pi=34.2atm$
