Question

what is the amount of NaOH present in 200 ml of 0.2N and aqueous solution?​

Answer 1

Answer: 1.6 g of NaOH / 0.04 moles of NaOH

Explanation:

Normality = no. of gram equivalent

—————————————

vol. of solvent. in litres

Given :

Volume of solvent = 200mL

= 0.2 L

Gram equivalent mass of NaOH = 23 + 16 + 1

= 40g

Normality = 0.2

Hence,

0.2 = x/40

——-

0.2

=> 0.2 x 0.2 = x/40

=> 0.2 x 0.2 x 40 = x

=> x = 1.6g

Hence 1.6 g of NaOH is present

= 1.6/40 moles of NaOH

=. 0.04 moles

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