Question

50mL of 0.5M Oxalic Acid is needed to neutralize 25mL of sodium hydroxide solution. The amount of NaOH in 50mL of the given sodium hydroxide solution is?


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Answer 1

Answer:

H 2C

2O4+2NaOH→Na

2C

2O

4+2H

2Om

eq of H 2C2O4 =m eq of NaOH

50×0.5×2=25×N

NaOH

×1

∴M

NaOH=N

NaOH =2 M ; since n-factor is 1

Now, 1000 ml solution =2×40 gram NaOH

∴50 ml solution =4 gram NaOH.

Explanation:

hope help uh

bohot lamba tha h na (〇o〇；)

Answer 2

Answer:4gm.

umber of gram equivalents of H2C2O4= Number of gram equivalents of NaOH

⇒50×0.5×2=25×x×1⇒x=50×0.5×225=2M

M=2=W40×100050⇒W=2×40×501000=4gm.

hope it helps u !!