Answer:
Concentration of [tex]H_3PO_4[/tex] in sample is 0.25 M.
Explanation:
From the reaction, one mole of [tex]H_3PO_4[/tex] reacts with 3 moles of NaOH.
Now, number of moles of NaOH, n = [tex]molarity \times volume(in \ liters).[/tex]
[tex]n=0.11\times \dfrac{24.83}{1000}\ mol=2.73\times 10^{-3}\ mol.[/tex]
Therefore, [tex]2.73\times 10^{-3}[/tex] mol of NaOH reacts with [tex]3\times 2.73\times 10^{-3}[/tex] [tex]H_3PO_4[/tex].
So, concentration of [tex]H_3PO_4[/tex] [tex]=\dfrac{no\ of \ moles}{volume\ in\ liter}=\dfrac{3\times 2.73\times 10^{-3}}{\dfrac{32}{1000}}=\dfrac{3\times 2.73}{32}=0.25\ M.[/tex]
Therefore, concentration of [tex]H_3PO_4[/tex] in sample is 0.25 M.
Hence, this is the required solution.
