Question

A $200\,ml\,{{I}_{2}}$ solution is divided into two unspecified sections. Part I reacts with hypo solution in acidic medium and requires 8ml of 2M hypo solution for complete neutralization.
Part II was added with 300ml of 0.1M NaOH solution. For complete neutralisation the residual base required 30ml of 0.1 M ${{H}_{2}}S{{O}_{4}}$ solution. Calculate the value of 20 times the initial concentration of ${{I}_{2}}$?
A.1
B.2
C.3
D.4

Answer 1

Hint: Hypo solution is the abbreviation for sodium thiosulphate or sodium hyposulfite, (whose chemical formula is $N{{a}_{2}}{{S}_{2}}{{O}_{3}}$ a chemical used since it was formed to bind the image to photographic film. It is often referred to as a "fixer," because it is used when processing them to extract residual silver halides from the films and prints.

Complete answer:
Molarity is defined as the number of moles of solute dissolved in one litre of the solution. It is denoted by the letter M.
N-factor for acids, n-factor is defined as the number of ${{H}^{+}}$ions replaced by 1 mole of acid in a reaction. For bases, it is defined as the number of $O{{H}^{-}}$ ions replaced by 1 mole of base in a reaction.
Part one:
Millimoles of $N{{a}_{2}}{{S}_{2}}{{O}_{3}}$ used =8×2×1( n-factor)=16
${{I}_{2}}+2N{{a}_{2}}{{S}_{2}}{{O}_{3}}\to 2NaI+N{{a}_{2}}{{S}_{4}}{{O}_{6}}$
1 mol 2 mol
Millimoles of ${{I}_{2}}$​ used =21​ Millimoles of $N{{a}_{2}}{{S}_{2}}{{O}_{3}}$​=216​=18
Part Two:
$3{{I}_{2}}+6NaOH\to 5NaI+NaI{{O}_{3}}+3{{H}_{2}}O$
Millimoles of ${{H}_{2}}S{{O}_{4}}$ ​= excess $NaOH$ =30×0.1×2=6
Millimoles of total $NaOH$ =300×0.1×1=30
Millimoles of $NaOH$ used =30−6=24
Using ${{I}_{2}}$​ millimoles = 21 millimoles of $NaOH$ = 224=12 mmol of ${{I}_{2}}$​
Total mmol used for ${{I}_{2}}$​ = Part I + Part II = 8 + 12=20 mmol
Molarity of ${{I}_{2}}$​ ​=$\dfrac{mmol}{VmL}$ =20020​=0.1 M
20 times the initial $M{{I}_{2}}$ =0.1×20=2

Hence the correct answer is option B

Note:
Moles can be measured as grammes or litres. One mole (mol) equals one thousand millimoles. Multiply by 1000 to convert moles to millimoles; divide by 1000 to convert millimoles to moles.
Molarity can be calculating by the formula:
$M\text{ }=\dfrac{n}{v}$
$M$= concentration in molars
$n$= Solute Moles
$v$= solution in litres