MHT CET Chemistry question

of oxalic acid completely neutralised 0.064 g of sodium hydroxide. Molarity of the oxalic acid solution is

Answer: (D)

Solution

Moles of NaOH = 0.064/40 = 0.0016 mol; oxalic acid is dibasic, so moles of oxalic acid = 0.0008 mol in 0.025 L, giving a

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Question ID VK-061186