Calculate the solubility (in text{moles/litre | vedclass

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(C) The relative lowering of vapour pressure is given by $\frac{P^0 - P_s}{P^0} = i 	imes X_{solute}$.For $Ag_3PO_4$,the van't Hoff factor $i = 4$ $(

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$\frac{10}{54}$

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(C) The relative lowering of vapour pressure is given by $\frac{P^0 - P_s}{P^0} = i 	imes X_{solute}$.For $Ag_3PO_4$,the van't Hoff factor $i = 4$ $(Ag_3PO_4 ightarrow 3Ag^+ + PO_4^{3-})$.$\frac{760 - 750}{760} = 4 	imes X_{solute}$.$\frac{10}{760} = 4 	imes X_{solute} \Rightarrow X_{solute} = \frac{1}{304}$.Since $X_{solute} = \frac{n_{solute}}{n_{solute} + n_{solvent}} \approx \frac{n_{solute}}{n_{solvent}}$,we have $\frac{n_{solute}}{n_{solvent}} = \frac{1}{304}$.For $1 \ L$ of water,$n_{solvent} = \frac{1000 \ g}{18 \ g/mol} = 55.55 \ mol$.$n_{solute} = \frac{55.55}{304} \approx 0.1827 \ mol/L$.$0.1827 \approx \frac{10}{54}$.

Calculate the solubility (in $\text{moles/litre}$) of a saturated aqueous solution of $Ag_3PO_4$ if the vapour pressure of the solution becomes $750 \ torr$ at $373 \ K$. (Assume vapour pressure of pure water at $373 \ K$ is $760 \ torr$ and density of water is $1 \ g/mL$)

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