Question

le 1 Calculate the normality of 10% solution of NaOH,​

Answer 1

Answer:

1.67

Explanation:

1.67

Answer 2

To Find -

• The Normality of 10 % solution of NaOH .

Concepts/ Formulae Involved -

• The concept of molarity and normality is used here.

• $Molarity =$ $\frac{Number\ of\ moles\ of\ solute (NaOH)}{Number\ of\ litres\ of\ solution}$

• $Normality = Molarity\ *\ N-Factor$

• Where, N-Factor for an acid is its basicity and N-Factor for a base is its acidity.

Value Used -

=> We use the density of 10 % (w/w) NaOH solution to be 1.11 g cm-3 at 20°C . (from SigmaAldrich website)

Let's Begin To Solve -

» For Molarity -

‣ Let the mass of the solution be 100 grams.

=> So, 10 % of this is NaOH= $\frac{10}{100}\ *\ 100\ grams$ => 10 grams is NaOH.

‣ Molar mass of NaOH = 23 + 16 + 1 => 40 grams per mole (Na=23,O=16,H=1)

=> So, 10 grams of NaOH is  $\frac{10}{40}\ =>\ \frac{1}{4}$ => 0.25 moles of NaOH .

‣ Volume of 100 grams of 10 % NaOH = ?

=> The density of 10 % NaOH solution = 1.11\ g\ *\ cm^{-3} at 20 °C .

∵ $Density\ =\ \frac{Mass}{Volume}$

=> 1.11 = $\frac{100}{Volume}$

=> Volume = $\frac{100}{1.11}$ => 90.09 mL

∴ Volume of solution in Litres = 90.09/1000 => 0.09009 L

‣ Molarity = $\frac{Number\ of\ moles\ of\ solute (NaOH)}{Number\ of\ litres\ of\ solution}$

=> Molarity = $\frac{0.25}{0.09009} => 2.775 M(molar)$  

=> Molarity = 2.775 M (molar)

» For Normality -

‣ Acidity of a base = number of OH-1 ions that are formed on complete dissociation from one molecule of it in aqueous state.

=> Formula = NaOH

=> $NaOH$ ⇌  $Na^{+}\ +\ OH^{-1}$ (In aqueous state)

=> Acidity of NaOH = 1

=> N-Factor of NaOH = 1

‣ Normality = Molarity * N-Factor

=> Normality = 2.775 * 1

=> Normality = 2.775 N

Answer -

» The Normality of 10 % NaOH solution is 2.775 N .

More to refer to -

>> For more information about interconversion of molarity and normality -

https://brainly.in/question/35760545