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**Hint:**We know that the number of gram equivalents in a solution remains the same.

i.e. ${N_1}{V_1} = {N_2}{V_2}$ The volume and concentration of the initial solution is given. The concentration of the second solution is given. (decinormal = 0.1N).

Therefore, find the volume of the second solution.

**Formula used:**

Note that, the number of gram equivalents in a solution remains the same.

i.e. ${N_1}{V_1} = {N_2}{V_2}$

Where,${N_1}$ = concentration of the first solution

${N_2}$ = concentration of the second solution

${V_1}$ = volume of the first solution

${V_2}$ = volume of the first solution

**Complete step by step answer:**

The concentration of a decinormal solution is given by $\dfrac{1}{{10}}$(N) i.e. 0.1 (N)

We know that the number of gram equivalents in a solution remains the same.

i.e. ${N_1}{V_1} = {N_2}{V_2}$

Where,${N_1}$ = concentration of the first solution = 10 (N)

${N_2}$ = concentration of the second solution = $\dfrac{1}{{10}}$(N) = 0.1(N)

${V_1}$ = volume of the first solution = 10 ml

${V_2}$ = volume of the first solution = ?

Now substituting the values in the equation ${N_1}{V_1} = {N_2}{V_2}$, we get

$10 \times 10 = 0.1 \times {V_2}$

$ \Rightarrow {V_2} = \dfrac{{100}}{{0.1}}$

$ \Rightarrow {V_2} = 1000{\text{ ml}}$

So the final volume of the solution will be 1000 ml.

Hence, (1000−10)=990 ml water should be added to the solution in order to make it decinormal.

**Therefore, the correct answer is option (A).**

**Note:**The concentration of a decinormal solution is given by $\dfrac{1}{{10}}$(N) i.e. 0.1 (N)

Find the final volume of the decinormal solution by substituting the values of ${N_1}$, ${N_2}$ , ${V_1}$ and ${V_2}$ in the equation ${N_1}{V_1} = {N_2}{V_2}$

Hence we can find the excess water needed by subtracting the initial volume from the final one.

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