Question

4 g of solid NaOH was dissolved in water and the
solution was made up to 1000 ml. The whole of
this will neutralized completely
(1) 100 ml of 1 M H.SO,
(2) 20 ml of 2.5 M H SO
(3) 20 ml of 1.5 M H,SO
(4) 30 ml of 5 M H,SO​

Answer 1

answer : option (2)

explanation : mass of solid NaOH = 4g

molecular mass of NaOH = 40g/mol

so, number of mole of NaOH = 4/40 =0.1

molarity = number of mole of solute/volume of solution in litre

= 0.1 × 1000/1000

= 0.1M

then, normality = n-factor × molarity

= 1 × 0.1 = 0.1N

for completely neutralisation,

equivalents of acid = equivalents of base

or, n- factor of acid × mole of acid = 1 × 0.1 = 0.1

[for H2SO4 , n -factor = 2 ]

or, mole of acid = 0.1/2 = 0.05

now checks all options ,

option (1) ⇒ mole of acid = 1 × 100/1000 = 0.1

option (2) ⇒mole of acid = 2.5 × 20/1000 = 50/1000 = 0.05 hence, option (2) is satisfied the above condition. so it is correct answer.

Answer 2

Answer:

option 2 is correct answer for this