$0,5g$ of benzoic acid is dissolved in $500mL75C$ hot water. This solution is divided into four aliquots, Each flask is placed into the thermostat and the temperature is adjusted as indicated in the table and wait until they reach thermal equilibrium. $100mL$ of each portion is taken into another flask and phenolphthalein indicator is added to each of them. Then they are titrated with $0\mathrm{.}05$MNaOH solution. By using the given information calculate the heat of solution of benzoic acid $($Molar mass BA$=122\mathrm{.}12$ g $($:$mo{l}^{-1},$NaOH$=39\mathrm{.}99(g)mo{l}^{-1},R=8\mathrm{.}314(J)K^{-1}mo{l}^{-1}\}$

a$)$ Calculate the solubility of benzoic acid in $100g$ water at each temperature

b$)$ Calculate the molar heat of solution of benzoic acid.

$ln(9\mathrm{.}992{10}^3)=-4\mathrm{.}60$

$\frac{1}{2}a8=3\mathrm{.}355kk{j}^{-3}$

$h(0\mathrm{.}015)=-4\mathrm{.}14$

$\frac{1}{313}=3\mathrm{.}196*{10}^{-3}$

$ln(0\mathrm{.}0139)=-3\mathrm{.}91$

$\frac{1}{323}={3\mathrm{.}29566}^3$

$ln(0\mathrm{.}2250,)=-765$

$\frac{1}{330}*3\mathrm{.}003{10}^{-3}$