Question
$$250\,mL$$ of sodium carbonate solution contains $$2.65\,g$$ of $$N{a_2}C{O_3}.$$ If $$10\,mL$$ of this solution is diluted to $$500\,mL,$$ the concentration of the diluted acid will be
A.
0.01$$\,M$$
B.
0.001$$\,M$$
C.
0.05$$\,M$$
D.
0.002$$\,M$$
Answer :
0.002$$\,M$$
Solution :
$$\eqalign{
& {\text{Molarity}}\,\,{\text{of}}\,\,N{a_2}C{O_3}\,\,{\text{solution}} \cr
& = \frac{{2.65}}{{106}} \times \frac{{1000}}{{250}} \cr
& = 0.1\,M \cr
& 10\,mL\,\,{\text{of this solution is diluted to}}\,\,500\,mL. \cr
& {M_1}{V_1} = {M_2}{V_2} \cr
& 0.1 \times 10 = {M_2} \times 500 \Rightarrow {M_2} = 0.002\,M \cr} $$