Question

How much water is to be added to dilute 10mL of 10N HCL to make it decinormal?
A. 990mL
B. 1010mL
C. 100mL
D. 1000mL

Answer 1

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.