Question

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

(1) 8g 

(2) 4g 

(3) 1g 

(4) 2g

Answer 1

The correct option (2) 4 g   

Explanation:

Mili equivalents of oxalic acid = mili equivalent of sodium hydroxide 

⇒ 50 × 0.5 × 2 = 25 × M × 1 

⇒ M = 2 molar

Mass of NaOH in 50ml solution 

= 50/1000 x 2 x 40 = 4g