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# For 1molar solution of $NaCl$ in water at $25^\circ C$ and $1$atm pressure show that:A. Molarity $=$ mole fractionB. Molality $=$ mole fractionC. Normality $=$ mole fractionD. Molarity $=$ normality

Hint:Molar mass is the mass of one mole of a substance. Equivalent weight of a substance (oxidizing and reducing agents) is equal to the molar mass divided by the number of gain or loss by one molecule of the substance in the redox reaction.
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.

Thus the correct answer is option (D).

Note:For 1 molar solution of $NaCl$ in water at $25^\circ C$ and $1$atm pressure show that normality becomes equal to molarity. For an acid, n -factor is defined as the number of ${H^ + }$ions replaced by$1$mole of acid in a reaction. But the n-factor for acid is not equal to its basicity. For example n-factor of sulphuric acid is $2.$ For a base, n-factor is the number of hydroxyl ions replaced by $1$ mole of a base in a reaction. But n-factor for base is not equal to its acidity. For example n-factor of sodium hydroxide is $1.$