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A. Molarity $ = $ mole fraction

B. Molality $ = $ mole fraction

C. Normality $ = $ mole fraction

D. Molarity $ = $ normality

Answer
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Formula used: Equivalent weight of an oxidizing agent $ = \dfrac{{Molar\,mass}}{{number\,of\,electrons\,\,gain\,\,{\text{b}}y\,\,metal}}$.

Given: \[1\] Molar$NaCl$means \[1\] moles of solute dissolved per litre of the solution

i.e. molarity of $NaCl$ is\[1.\]

In the given $NaCl,$charge on sodium and chloride ion is\[1\,\,(\because \,NaCl \to N\mathop a\limits^ + + C\mathop l\limits^ - )\]. Where n-factor is the total charge on positive ion or total charge on negative ion.

Thus, n-factor of$NaCl$is\[1.\]

We know that normality is defined as the number of gram equivalent of solute dissolved per liter of the solution.

Therefore Normality ${\text{ = }}\dfrac{{{\text{number of gram equivalent of solute}}}}{{{\text{volume of solution(ml)}}}} \times 1000\,$……. (i)

Molarity is defined as the number of moles of solute dissolved per litre of the solution.

Therefore Molarity $ = {\text{ }}\dfrac{{{\text{number of moles of solute}}}}{{{\text{volume of solution(ml)}}}} \times 1000$………. (ii)

From equation (i) and (ii)

Divide (ii) from (i), we get

$\Rightarrow \dfrac{{{\text{Normality}}}}{{Molarity}} = \dfrac{{number{\text{ of gram equivalent of solute}}}}{{number\,of\,moles\,of\,solute}}$

$\Rightarrow \dfrac{{{\text{Normality}}}}{{Molarity}} = \dfrac{{\dfrac{{w{\text{eight of solute}}}}{{Equivalent\,weight}}}}{{\dfrac{{{\text{Weight of solute}}}}{{molar\,mass}}}}{\text{ }}$

(Where, number of gram equivalent of solute $ = \,\,\dfrac{{weight}}{{Equivalent{\text{ weight}}}}$ and

Number of moles of solute\[ = \dfrac{{weight}}{{Molar{\text{ mass}}}}\])

$\Rightarrow \dfrac{{{\text{Normality}}}}{{Molarity}} = \dfrac{{Molar\,mass}}{{\dfrac{{{\text{Molar mass}}}}{{n - factor}}}}$ ($\because $ Equivalent weight$\, = {\text{ }}\dfrac{{{\text{Molar mass}}}}{{n - factor}}$)

$\dfrac{{{\text{Normality}}}}{{Molarity}} = $ ${\text{n - factor}}$

$\therefore $ Normality $ = $ Molarity $ \times $ n-factor

Now, putting the value of molarity of $NaCl$ and n-factor of $NaCl.$

$ \Rightarrow $ Normality $ = $ $1 \times 1$

$\therefore $ Normality $ = \,\,1$

Hence, the normality is equal to molarity.