Chemistry - Mole Concept Question with Solution | TestHub

$500ml0.2MN{a}_{2}S{O}_4$ solution

Moles of $N{a}_{2}S{O}_4=\frac{500}{1000}\times 0.2=0.1$ mole

Moles of $N{a}^{+}=0.2$ mole

Moles of $S{O}_4^{2-}=0.1$ mole

$\Rightarrow 17.1\left(\frac{W}{V}\right)\%A{l}_2(S{O}_4{)}_{3}in100ml$

means $17.1g$ of $A{l}_2(S{O}_4{)}_3$ in $100ml$ solution.

$$\begin{gathered} \text{displaylines}=\frac{17.1}{342}\times \frac{1}{10} \\ =\frac{1}{2}\times \frac{1}{10}=\frac{1}{20} \end{gathered}$$

Moles of $A{l}^{3+}=2\times \frac{1}{20}=\frac{1}{10}$

Moles of $S{O}_4^{3-}=3\times \frac{1}{20}$

Total moles of $$\begin{gathered} \text{displaylines}S{O}_4^{3-}=0.1+\frac{3}{20} \\ =\frac{1}{10}+\frac{3}{20} \\ =\frac{2+3}{20}=\frac{5}{20}=\frac{1}{4} \end{gathered}$$

Total volume of solution $=500ml+100ml=600ml=0.6L$.

on 5 times dilution volume

$$\begin{gathered} \text{displaylines}=600\times 5 \\ =3000ml=3L \end{gathered}$$

Molarity of $S{O}_4^{2-}=\frac{Totalmoles}{Totalvolume}=\frac{\frac{1}{4}}{3}=\frac{1}{12}M$