Question

If 8.3 ml of a sample of \[{H_2}S{O_4}\] (36 N) is diluted by 991.7 ml of water, the approximate normality of the resulting solution is:
A. 0.4
B. 0.2
C. 0.1
D. 0.3

Answer 1

Hint- In order to solve the problem first we will find the volume of sample and the solution for both the cases. We will use the relation between the normality of a solution and the volume of the solution in order to find the answer. The product of normality of sample and the volume of the sample remains constant.

Complete answer:
Formula used- ${N_1} \times {V_1} = {N_2} \times {V_2}$
At initial stage:
Normality of \[{H_2}S{O_4}\] sample $\left( {{N_1}} \right) = 36N$
Volume of \[{H_2}S{O_4}\] sample $\left( {{V_1}} \right) = 8.3ml$
Volume of the water added $ = 991.7ml$
So, total volume of the solution = volume of \[{H_2}S{O_4}\] sample + volume of water added
$
  \left( {{V_2}} \right) = 8.3ml + 991.7ml \\
  \left( {{V_2}} \right) = 1000ml \\
 $
Now let us calculate the normality of the resulting solution $\left( {{N_2}} \right)$
As we know that the product of normality of sample and the volume of the sample remains constant.
$ \Rightarrow {N_1} \times {V_1} = {N_2} \times {V_2}$
Let us substitute all the values in the equation:
$
  \because {N_1} \times {V_1} = {N_2} \times {V_2} \\
   \Rightarrow 36N \times 8.3ml = {N_2} \times 1000ml \\
 $
Now let us solve the above equation for the value of $\left( {{N_2}} \right)$
$
   \Rightarrow {N_2} = \dfrac{{36N \times 8.3ml}}{{1000ml}} \\
   \Rightarrow {N_2} = \dfrac{{298.8N - ml}}{{1000ml}} \\
   \Rightarrow {N_2} = 0.2988N \\
   \Rightarrow {N_2} \approx 0.3N \\
 $
Hence, the approximate normality of the resulting solution is 0.3.
So, option D is the correct option.

Note- Normality is described as the number of gram or mole equivalents of solute present in one liter of a solution. When we say equivalent, it is the number of moles of reactive units in a compound. Normality is used to measure the concentration of a solution. It is mainly used as a measure of reactive species in a solution and during titration reactions or particularly in situations involving acid-base chemistry.