50   mL of 0 .5   M oxalic acid is needed to neutralize 25   mL of sodium hydroxide…

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?
  1. 2 g
  2. 4 g
  3. 1 g
  4. 8 g

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=80g NaOH

∴ 50ml NaOH solution contains =801000×50=4 g

Asked in: JEE Main 2019 (12 Jan Shift 1)

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