Question
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 (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
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
At initial stage:
Normality of H2SO4 sample (N1)=36N
Volume of H2SO4 sample (V1)=8.3ml
Volume of the water added =991.7ml
So, total volume of the solution = volume of H2SO4 sample + volume of water added
(V2)=8.3ml+991.7ml (V2)=1000ml
Now let us calculate the normality of the resulting solution (N2)
As we know that the product of normality of sample and the volume of the sample remains constant.
⇒N1×V1=N2×V2
Let us substitute all the values in the equation:
∵N1×V1=N2×V2 ⇒36N×8.3ml=N2×1000ml
Now let us solve the above equation for the value of (N2)
⇒N2=1000ml36N×8.3ml ⇒N2=1000ml298.8N−ml ⇒N2=0.2988N ⇒N2≈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.