Solveeit Logo
Login

Question

Question: For 1molar solution of \[NaCl\] in water at \(25^\circ C\) and \[1\]atm pressure show that: A. Mol...

For 1molar solution of NaClNaCl in water at 25∘C25^\circ C and 11atm pressure show that:
A. Molarity == mole fraction
B. Molality == mole fraction
C. Normality == mole fraction
D. Molarity == normality

Explanation

Solution

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 =Molar massnumber of electrons  gain  by  metal = \dfrac{{Molar\,mass}}{{number\,of\,electrons\,\,gain\,\,{\text{b}}y\,\,metal}}.

Complete answer:
Given: 11 MolarNaClNaClmeans 11 moles of solute dissolved per litre of the solution
i.e. molarity of NaClNaCl is1.1.
In the given NaCl,NaCl,charge on sodium and chloride ion is1  (∵ NaCl→Na++Cl−)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 ofNaClNaClis1.1.
We know that normality is defined as the number of gram equivalent of solute dissolved per liter of the solution.
Therefore Normality  = number of gram equivalent of solutevolume of solution(ml)×1000 {\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 = number of moles of solutevolume of solution(ml)×1000 = {\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
⇒NormalityMolarity=number of gram equivalent of solutenumber of moles of solute\Rightarrow \dfrac{{{\text{Normality}}}}{{Molarity}} = \dfrac{{number{\text{ of gram equivalent of solute}}}}{{number\,of\,moles\,of\,solute}}
⇒NormalityMolarity=weight of soluteEquivalent weightWeight of solutemolar mass \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 =  weightEquivalent weight = \,\,\dfrac{{weight}}{{Equivalent{\text{ weight}}}} and
Number of moles of solute=weightMolar mass = \dfrac{{weight}}{{Molar{\text{ mass}}}})

⇒NormalityMolarity=Molar massMolar massn−factor\Rightarrow \dfrac{{{\text{Normality}}}}{{Molarity}} = \dfrac{{Molar\,mass}}{{\dfrac{{{\text{Molar mass}}}}{{n - factor}}}} (∵\because Equivalent weight = Molar massn−factor\, = {\text{ }}\dfrac{{{\text{Molar mass}}}}{{n - factor}})
NormalityMolarity=\dfrac{{{\text{Normality}}}}{{Molarity}} = n - factor{\text{n - factor}}
∴\therefore Normality == Molarity ×\times n-factor
Now, putting the value of molarity of NaClNaCl and n-factor of NaCl.NaCl.
⇒\Rightarrow Normality == 1×11 \times 1
∴\therefore Normality =  1 = \,\,1
Hence, the normality is equal to molarity.

Thus the correct answer is option (D).

Note: For 1 molar solution of NaClNaCl in water at 25∘C25^\circ C and 11atm pressure show that normality becomes equal to molarity. For an acid, n -factor is defined as the number of H+{H^ + }ions replaced by11mole 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 2.2. For a base, n-factor is the number of hydroxyl ions replaced by 11 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 1.1.