Question

0.115 g of pure sodium metal was dissolved in 500 ml distilled water. The normality of the above solution, whose resulting volume is 400 ml, would be;
A. 0.010 N
B. 0.0115 N
C. 0.0125 N
D. 0.046 N

Answer 1

Hint: Normality of a solution is the number of gram equivalents of the solute dissolved per litre of a given solution. It is signified by N.

Formula Used:
\[{\rm{N=}}\dfrac{{{\rm{no}}{\rm{. \ of \ gram \ equivalent \ of \ solute}}}}{{{\rm{Volume \ of the \ solution \ in \ litres}}}}\].

Complete Step by Step Solution:
Normality is the ratio of no.of gram equivalents of the solute dissolved per litre of a given solution.
If 'w' gram of the solute is present in V litres of a given solution,
\[{\rm{N=}}\left( {\dfrac{{\rm{w}}}{{{\rm{eq}}{\rm{.mass \ of \ the \ solute}}}}} \right){\rm{ \times }}\left( {\dfrac{{{\rm{1000}}}}{{\rm{V}}}} \right)\]
Here, we are given the mass of pure sodium metal, 0.115g.
The volume of the given solution = 500mL.
Equivalent mass of sodium metal in the solution
\[ = \left( {\dfrac{{{\rm{atomic \ mass}}}}{{{\rm{valency}}}}} \right)\]

The atomic mass of Sodium is 23g. The valency of Sodium is 1.
So, the equivalent mass of Sodium
= 23 g/equivalent = 23 amu (atomic mass unit)

So, the normality of Sodium metal in the given solution
$=\left ( \dfrac{0.115 \ g}{23 \ amu} \right )\times\left (\dfrac{1000}{500 \ L} \right )$
$= 0.010 N $
So, option A is correct.

Additional Information:Molarity and Normality comprises volumes of solutions. The temperature has a significant effect on volumes. These values change with temperature. So, molality, mole fraction, mass fraction, etc. have more preference as these depend on masses. Mass does not change with temperature.

Note: While attending to the question, one must have remembered the formula of normality. The equivalent mass of a solute formula also has to be remembered. Units of different values have to be mentioned during calculations.