Question
What volume of $$5\,M\,N{a_2}S{O_4}$$ must be added to $$25\,mL$$ of $$1\,M\,BaC{l_2}$$ to produce $$10\,g$$ of $$BaS{O_4}?$$
A.
8.58$$\,mL$$
B.
7.2$$\,mL$$
C.
10$$\,mL$$
D.
12$$\,mL$$
Answer :
8.58$$\,mL$$
Solution :
$$N{a_2}S{O_4} + BaC{l_2} \to BaS{O_4} + 2NaCl$$
$${\text{No}}{\text{. of moles of}}\,\,BaS{O_4}$$ $$ = \frac{w}{M} = \frac{{10}}{{233}} = 0.0429$$
$$\therefore \,\,{\text{No}}{\text{. of moles of}}\,\,N{a_2}S{O_4}\,\,{\text{needed}}$$ $$ = \frac{{M \times V}}{{1000}}$$
$${\text{or}}\,\,\,0.0429 = \frac{{5 \times V}}{{1000}} \Rightarrow V = 8.58\,mL$$