50   mL of 0 .5   M oxalic acid is needed to neutralize 25   mL of sodium hydroxide solution. What is the amount of NaOH in 50   mL of the given sodium hydroxide solution?

Step-by-step solution

millimoles of oxalic acid = 50 × 0.5 milli equivalent of oxalic acid = 50 × 0.5 × 2 millimoles of NaOH = 25 × M = milli equivalent of NaOH ∴ 50 × 0 .5 × 2 = 25 × M ⇒ M = 2 M = 2 so, 1000   ml NaOH solution contains = 2  mol of  NaOH = 2 × 40 = 80 g  NaOH ∴   50 ml  NaOH solution contains = 80 1000 × 50 = 4   g