Question

50 mL of 0.5 M oxalic acid is needed to neutralize 25 mL of sodium hydroxide solution. The amount of NaOH in 50 mL of the given sodium hydroxide solution is : [JEE(Main)-2019, January]
(1) 40 g (2) 20 g (3) 80 g (4) 10 g​

Answer 1

Explanation:

Given 50 mL of 0.5 M oxalic acid is needed to neutralize 25 mL of sodium hydroxide solution. The amount of NaOH in 50 mL of the given sodium hydroxide solution is :

The reaction will be milli equivalent of oxalic acid = milli equivalent of sodium hydroxide.

Now 50 x 0.5 x 2 = 25 x M naoh x 1

  So M NaOh = 2

Now 1000 ml solution contains = 2 x 40 g = 80 g of NaOh

Therefore 50 ml solution contains 80 x 50 / 1000

                                                  = 4g NaOh