Question

4.0 gm of NaOH is dissolved in 100 cc of solution. The normality of the solution is:
(A) 1.0
(B) 0.1
(C) 0.5
(D) 0.4

Answer 1

Hint: The given problem can be solved by knowing the basics of normality and the equations based on the same. Also, from the given options, only one option can be correct.

Complete step by step solution:
Let us solve the given illustration directly, we just need to know the basics of normality as explained below;
Normality- Normality is the number of gram equivalents or mole equivalents of solute present in one litre of a solution.
Mole equivalent is the number of moles who are actually the reactive units in the reaction. Similar goes with the gram equivalent.
It is expressed as,
\[N=M\times n\]
where, n is the ratio of molar mass to the equivalent mass of the component.
Equivalent weight- Equivalent weight can be expressed in terms of acidity for bases ($O{{H}^{-}}$) and basicity for acids (${{H}^{+}}$). This can be given as,
$Eq.wt.=\dfrac{Mol.wt.}{Acidity}$ and $Eq.wt.=\dfrac{Mol.wt.}{Basicity}$
Also,
Normality can simply be expressed as,
\[N=\dfrac{{{n}_{g}}}{V}\]
where,
V = volume of solution in litres
${{n}_{g}}$ = number of gram equivalents i.e. it is the ratio of given mass to the equivalent mass of a compound.
Now, let us go towards the illustration,
Given that,
Given mass of NaOH = 4 gm
Volume of solution = 100 cc = 0.1 L
Molar mass of NaOH = 40 gm/mol
Equivalent mass can be calculated as,
$NaOH\to N{{a}^{+}}+O{{H}^{-}}$
$Eq.wt.=\dfrac{Mol.wt.}{Acidity}=\dfrac{40}{1}=40gmmo{{l}^{-1}}e{{q}^{-1}}$
Thus,
Equivalent moles can be given as,
$Eq.moles=\dfrac{4gm}{40gmmo{{l}^{-1}}e{{q}^{-1}}}=0.1$
Now, normality can be calculated as,
$Normality=\dfrac{0.1}{0.1}=1N$

Therefore, option (A) is correct.

Note: Do note to use the units properly and do not get confused in concepts of molarity and normality. Though they are interrelated, their multiplying factors differ.