Question

What amount of oxalic acid is required to prepare $250{\text{ }}ml$. of $0.1N$ solution?
A) $1.275g$
B) $5.575{\text{ }}g$
C) $1.575{\text{ }}g$
D) $1.550g$

Answer 1

Normality: The number of gram equivalents of a substance present in a litre of solution is called normality.
In current practice, concentration is most often expressed as molarity. Molarity is defined as
the number of moles of solute per litre of solution.
Molarity$= \dfrac{{moles\,\,of\,\,solute}}{{volume\,\,(solution)\;in\,\,litres}}$ or $M = \dfrac{{w/M}}{{V/1000}} = \dfrac{{w \times 1000}}{{M'V}}$
w = Mass of solute in grams
M= molecular weight of solute in gm/mol.
V = volume of solution in ml.
$0.01{\text{ }}M{\text{ }}NaOH$ , solution means that 0.01 mole NaOH is present in $1000{\text{ }}ml$ of its solution.

Complete answer:
Normality $= \dfrac{{Number\,of\,gram - \,equivalents\,\,of\,\,solute}}{{Volume\,\,of\,\,solution\,\,in\,\,litres}}$
or $N = \dfrac{{w/E}}{{V/1000}} = \dfrac{{w \times 1000}}{{VE}}$
where w = mass of solute in gram
V = volume of solution in ml
E = equivalent wt of solute
Number of gram equivalents$= normality\;\, \times volume$
$= 0.1\, \times \,0.25\, = \,0.025$
Number of gram equivalents $= \dfrac{{weight}}{{equivalent\,weight}} = \dfrac{x}{{63}}$
$\Rightarrow 0.025 = \dfrac{x}{{63}}$
$\Rightarrow x = 0.025\, \times \,63\, = 1.575g$

**Therefore Option C. $1.575{\text{ }}g$ is the correct option.

Note:**
Equivalent weight: Equivalent weight of any substance is the weight of substance, which can combine, replace or replaced by $1g$ of hydrogen, $35.5g$ of chlorine or $8g$ of oxygen.
For example: $Ca + {H_2} \to {\text{ }}Ca{H_2}$ , In this reaction $40g$ Ca combines with $2{\text{ }}grams$ of hydrogen, so according to definition, the mass of Ca which combine with $1g$ hydrogen will be the equivalent weight of Ca and it is 20.