$0.756 g$ of oxalic acid dihydrate $H X_{2} C X_{2} O X_{4} \cdot 2 H X_{2} O$ (molar mass $126 g mo l^{- 1}$ ) is dissolved in water and made up to $100 mL$ . A $10.0 mL$ portion needs $12.5 mL$ of a $NaOH$ solution for complete neutralisation (to $Na X_{2} C X_{2} O X_{4}$ ). The molarity of the $NaOH$ is:

Answer: D

Moles of acid in

10.0 mL = 0.756/126 \times 0.1 = 0.00060

; with

n = 2

, equivalents

= 0.00120

. Moles of

NaOH

(

n = 1

) in

12.5 mL

equal this, so

M = 0.00120/0.0125 = 0.096 M