1.

25 cm^(3) of oxalic acid completely neutralised 0.064 g of sodium hydroxide. Molarity of the oxalic acidsolution is

Answer»

0.064
0.045
0.015
0.032

Solution :`undersetoverset(|)(COOH)(C )OOH + 2NAOH to undersetoverset(|)(COONA)(C ) OONa + 2H_2O`
MOLES of 0.064g of NaOH
` = (0.064)/(40) = 0.0016` mol
2 moles of NaOH react with 1 mol of oxalic acid
` therefore ` 0.0016 mol of NaOH react with oxalic acid
` = (0.0016)/(2) = 8 xx 10^(-4) ` mol
volume of solution = 25 mL
Molarity of oxalic acid solution` = (8 xx 10^(-4) xx 1000)/(25)`
0.032 M


Discussion

No Comment Found

Related InterviewSolutions