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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 nor...

If 8.3 ml of a sample of H2SO4{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

Explanation

Solution

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- N1×V1=N2×V2{N_1} \times {V_1} = {N_2} \times {V_2}
At initial stage:
Normality of H2SO4{H_2}S{O_4} sample (N1)=36N\left( {{N_1}} \right) = 36N
Volume of H2SO4{H_2}S{O_4} sample (V1)=8.3ml\left( {{V_1}} \right) = 8.3ml
Volume of the water added =991.7ml = 991.7ml
So, total volume of the solution = volume of H2SO4{H_2}S{O_4} sample + volume of water added
(V2)=8.3ml+991.7ml (V2)=1000ml  \left( {{V_2}} \right) = 8.3ml + 991.7ml \\\ \left( {{V_2}} \right) = 1000ml \\\
Now let us calculate the normality of the resulting solution (N2)\left( {{N_2}} \right)
As we know that the product of normality of sample and the volume of the sample remains constant.
N1×V1=N2×V2\Rightarrow {N_1} \times {V_1} = {N_2} \times {V_2}
Let us substitute all the values in the equation:
N1×V1=N2×V2 36N×8.3ml=N2×1000ml  \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 (N2)\left( {{N_2}} \right)
N2=36N×8.3ml1000ml N2=298.8Nml1000ml N2=0.2988N N20.3N  \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.