Question

what is the normality of 0.1381 M NaOH​

Answer 1

Answer:

Explanation:Normality=Molarity×{Mol. Wt/Eq. Wt)

Mol. Wt=23+16+1=40

Eq. Wt=40/2=20

Normality=(0.1381×40)/20

Normality= 0.2762N

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Answer 2

Answer:

The Normality of 0.1381 M NaOH is 5.524 N.

Explanation:

Normality can be defined as the mole equivalents or gram equivalents of solute present in one liter of the solvent.

So, $N = \frac{Gram Equivalent of the Solute}{Volume of the Solution in Liters}$

Gram Equivalent of the Solute is given by molar mass by equivalence factor, i.e.,

$Gram Equivalent of the Solute = \frac{Molar Mass}{Equivalence Factor}$

Also, Gram Equivalent of a Base can be given by, number of moles multiplied by valency factor, i.e.,

$Gram Equivalent of a Base = Number of Moles * Valency Factor$

Valency Factor of NaOH = 1

Molarity of NaOH = 0.1381 M = 0.1381 moles/L

And we know that  mol of NaOH = 40 g

So, $Gram Equivalent of NaOH = 0.1381 mol/L * 1 L * 40 g= 5.524 g$

Molarity can be defined as the amount of a substance present in one liter of solvent.

So, $M = \frac{Number of Moles of the Solute}{Volume of the Solvent in Liters}$

Hence, 0.1381 M means 0.1381 moles of NaOH is present in 1 Liter of Solvent.

So, the Volume = 1 Liter

Hence, $Normality of NaOH = 5.524 g/ 1 Liter= 5.524 N$

Thus, the Normality of 0.1381 M NaOH is 5.524 N.