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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 :

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4 g

10 g

20 g

80 g

answer is A.

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Detailed Solution

I H₂C₂O₄
V_a = 50ml
M_a = 0.5M
n_a = 2
NaOH
V_b = 25ml
n_b = 1

N_b = ?
V_aN_a = V_bN_b
50(0.5)2 = 25(N_b)
N_b = 2N
II V_{NaOH solution}= 50ml

\large N\; = \;\frac{w}{{GEW}} \times \frac{{1000}}{v}

\large 2\; = \;\frac{w}{{40}} \times \frac{{1000}}{{50}}

W_NaOH= 4g

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