Question

What volume of 0.2 M $KMn{O}_{4}$ is required to react with 1.58g of hypo solution $({Na}_{2}{S}_{2}{O}_{3})$ in acidic medium?
A.20 mL
B.10 mL
C.16.6 mL
D.50 mL

Answer 1

Hint:
Molarity is an expression that is used to calculate or measure the concentration of a solution. It can be used to calculate the volume of the solvent or the amount of solute.

Complete step by step answer:
Molarity (M) is defined as the number of moles of solute per litre of the solution. Now, let us look at the given question.
In an acidic medium, the following conversion takes place.
$\overset { +7 }{ Mn } { O }_{ 4 }^{ - }\quad \longrightarrow \quad { Mn }^{ +2 }$
In this, the oxidation of Mn changes from +7 to +2 and thus is reduced.
Also, the n-factor, ${n}_{1}$, of this reaction = 5
The following reaction also takes place.
$\overset { +2 }{ { S }_{ 2 } } { O }_{ 3 }^{ 2- }\quad \longrightarrow \quad \overset { +2.5 }{ { S }_{ 4 } } { O }_{ 6 }^{ 2- }$
In this reaction, the oxidation state of sulphur changes from +2 to +2.5 and thus is oxidized.
Also, the n-factor, ${n}_{2}$, of this reaction = 1

Now, let us calculate the no. of moles of ${Na}_{2}{S}_{2}{O}_{3}$
$No.\quad of\quad moles\quad of\quad { Na }_{ 2 }{ S }_{ 2 }{ O }_{ 3 }\quad =\quad \cfrac { Weight }{ Molecular\quad weight }$
The weight of ${Na}_{2}{S}_{2}{O}_{3}$ is given as 1.58g and the molecular weight of ${Na}_{2}{S}_{2}{O}_{3}$ is 158 amu. Substituting the values in the formula for no. of mole, we get,
$No.\quad of\quad moles\quad of\quad { Na }_{ 2 }{ S }_{ 2 }{ O }_{ 3 }\quad =\quad \cfrac { 1.58 }{ 158 }$
$\implies No.\quad of\quad moles\quad of\quad { Na }_{ 2 }{ S }_{ 2 }{ O }_{ 3 }\quad =\quad { 10 }^{ -2 }$

Now, applying the equivalence concept, we can find the volume of the $KMn{O}_{4}$ that is required.
$Equivalent\quad of\quad KMn{ O }_{ 4 }\quad =\quad Equivalent\quad of\quad { Na }_{ 2 }{ S }_{ 2 }{ O }_{ 3 }$
${ M }_{ 1 }\quad \times \quad { n }_{ 1 }\quad \times \quad { V }_{ 1 }\quad =\quad Mole\quad \times \quad { n }_{ 2 }$
Now, ${M}_{1}$ = 0.2M, ${n}_{1}$ = 5, ${n}_{2}$ = 1, Moles = ${10}^{-2}$. Substituting these values in the above equation, we get,
$0.2\quad \times \quad 5\quad \times \quad { V }_{ 1 }\quad =\quad { 10 }^{ -2 }\quad \times \quad 1$
$\implies { V }_{ 1 }\quad =\quad \cfrac { { 10 }^{ -2 } }{ 0.2\quad \times \quad 5 }$
$\implies { V }_{ 1 }\quad =\quad { 10 }^{ -2 }\quad L\quad =\quad 10\quad mL$
Therefore, the volume of 0.2 M $KMn{O}_{4}$ that is required to react with 1.58g of hypo solution $({Na}_{2}{S}_{2}{O}_{3})$ in acidic medium is 10 mL.
Hence, option (b) is the correct option.

Note:
Molarity and molality are two different quantities used to find the concentration of the
solution. Molarity (M) is defined as the number of moles of solute per liter of solution. And molality
(m) is defined as the number of moles of solute per kilogram of solvent. Make sure you do not
confuse them.