Question

Hlo friends!☺️
Calculate normality of solution containing 5g of NaOH dissolve in 250mL of aqueous solution.
Thanks for answering!☺️​

Answer 1

Solution:-

Moles of NaOH = 5/40 moles

=> Moles of NaOH = 1/8 Moles.

Molarity = (1/8)× 1000/250

=> Molarity = 1/2

Molarity = Normality * n factor.

Hence ,

Normality = 1/2.

Hope it helps!+

Answer 2

Given :

Mass of NaOH = 5 gm

Volume of the solution (V) = 250 ml

To Find :

Normality of the solution

Solution :

Normality of a solution is the number of gram-equivalents of a solute dissolved in one litre of the solution.

Normality (N) = Mass of the solute (W) ÷ [gram-equivalent mass (E) × Volume of solution (V)]

Normality (N) = $\frac{W}{E*V}$

Gram-equivalent weight of NaOH = $\frac{40}{1}$

                                                        = 40 gm

Normality (N) = $\frac{5*1000}{40 * 250}$

                      = 0.5 N

Therefore, the normality of the given solution is found to be 0.5 N