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50 mL of 0.5 M oxalic acid is needed to neutralise 25. mL of sodium hydroxide
solution. The amount of NaOH in 50 mL of the given sodium hydroxide solution is
(2019 Main, 12 Jan I)
A) 40 g
B) 80 g
C) 20 g
D) 10g​

Answer 1

Answer:

Given 50 mL of 0.5 M oxalic acid is needed to neutralize 25 mL of sodium hydroxide solution. The amount of NaOH in 50 mL of the given sodium hydroxide solution is : The reaction will be milli equivalent of oxalic acid = milli equivalent of sodium hydroxide.

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Answer 2

Answer:

ANSWER

H

2

C

2

O

4

+2NaOH→Na

2

C

2

O

4

+2H

2

O

m

eq

of H

2

C

2

O

4

=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:

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