Question

What is the equivalent weight of\[{\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}{\text{.5}}{{\text{H}}_{\text{2}}}{\text{O}}\]?
A) 240 g
B) 246 g
C) 248 g
D) 250 g

Answer 1

Hint: To calculate the equivalent weight of the compound \[{\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}{\text{.5}}{{\text{H}}_{\text{2}}}{\text{O}}\], the value of the molecular weight compound \[{\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}{\text{.5}}{{\text{H}}_{\text{2}}}{\text{O}}\]and the number of electrons transfer (loss/gain) by the compound are required.

Formula Used:
${\text{Molecular weight = }}\sum {{\text{atomic weight}}} $

${\text{Equivalent weight = }}\dfrac{{{\text{Molecular weight }}}}{{{\text{Number of electrons transferred}}}}$

Complete answer:
Using the atomic weights of all atoms in \[{\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}{\text{.5}}{{\text{H}}_{\text{2}}}{\text{O}}\]calculate the molecular weight of the compound.

${\text{Molecular weight = }}\sum {{\text{atomic weight}}} $
Atomic weight of \[{\text{Na}}\]= 23.0g
Atomic weight of \[{\text{S}}\]= 32.0g
Atomic weight of ${\text{O}}$ = 16.0g
Atomic weight of ${\text{H}}$ = 32.0g

Now, substitute the values of atomic weights and calculate the molecular weight of \[{\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}{\text{.5}}{{\text{H}}_{\text{2}}}{\text{O}}\]as follows:
In the chemical formula of \[{\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}{\text{.5}}{{\text{H}}_{\text{2}}}{\text{O}}\] there are 2 \[{\text{Na}}\]atoms, 2 \[{\text{S}}\] atoms, 8 ${\text{O}}$ atoms and 10 ${\text{H}}$ atoms present.

\[{\text{Molecular weight of N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}{\text{.5}}{{\text{H}}_{\text{2}}}{\text{O}} = 2(23.0{\text{g) + 2 (32}}{\text{.0 g) + 8(16}}{\text{.0 g) + 10(1}}{\text{.0g)}}\]
\[{\text{Molecular weight of N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}{\text{.5}}{{\text{H}}_{\text{2}}}{\text{O}} = 248{\text{ g}}\]

Using the standard redox reaction of \[{\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}{\text{.5}}{{\text{H}}_{\text{2}}}{\text{O}}\]with ${{\text{I}}_{\text{2}}}$ we can determine the number of electrons transferred.
The standard balance reaction \[{\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}{\text{.5}}{{\text{H}}_{\text{2}}}{\text{O}}\] with ${{\text{I}}_{\text{2}}}$ is
${\text{2N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}{\text{ + }}{{\text{I}}_{\text{2}}} \to {\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{4}}}{{\text{O}}_{\text{6}}}{\text{ + 2NaI}}$

Here, 2 moles of ${\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ reacts with 1 mole of ${{\text{I}}_{\text{2}}}$ and gives 2 moles of${\text{NaI}}$.

The oxidation number of elemental ${{\text{I}}_{\text{2}}}$ is zero while the oxidation number of ${\text{I}}$ in ${\text{NaI}}$ is -1. So, to form 2 moles of ${\text{NaI}}$ there is a transfer of 2 electrons.

Thus, we can conclude that when 2 moles of ${\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ reacts there is a loss of 2 electrons.

\[{\text{2 moles of N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}} = 2{\text{ electrons}}\]
\[{\text{1 moles of N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}} = 1{\text{ electron}}\]

Now, using the molecular weight of \[{\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}{\text{.5}}{{\text{H}}_{\text{2}}}{\text{O}}\] and the number of electrons loss calculate the equivalent weight of \[{\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}{\text{.5}}{{\text{H}}_{\text{2}}}{\text{O}}\] as follows:

${\text{Equivalent weight = }}\dfrac{{{\text{Molecular weight }}}}{{{\text{Number of electrons transferred}}}}$

${\text{Equivalent weight of N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}{\text{.5}}{{\text{H}}_{\text{2}}}{\text{O = }}\dfrac{{{\text{248 g }}}}{1} = 248{\text{ g}}$

Thus, the equivalent weight of \[{\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}{\text{.5}}{{\text{H}}_{\text{2}}}{\text{O}}\] is 248 g.

Hence, the correct answer is an option (C) 248 g.

Note:Equivalent weight of the compound depends on the number of electrons loss/gain. Use the oxidation number rules to determine the oxidation number of reacting species. The oxidation number of an element is always zero. The oxidation number of oxygen atom is always -2 except in peroxide it is -1. The oxidation number of a hydrogen atom is +1 except in metal hydride it is -1.