An aqueous solution of oxalic acid dihydrate contains its 6.3 g in 250 ml. The volume of 0.1 N NaOH required to completely neutralize 10 ml of this solution

Question

An aqueous solution of oxalic acid dihydrate contains its 6.3 g in 250 ml. The volume of 0.1 N NaOH required to completely neutralize 10 ml of this solution (A) 4ml (B) 20ml (C) 2ml (D) 40ml

Tardigrade Admin · Accepted Answer

Normality of oxalic add solution = (6.3×1000/63 × 250) = 0.4 N Now from N1V1 = N2V2 0.4 × 10 = 0.1 × V2 V2= 40 mL

Q. An aqueous solution of oxalic acid dihydrate contains its 6.3 g in 250 ml. The volume of 0.1 N NaOH required to completely neutralize 10 ml of this solution

4ml

20ml

2ml

40ml

Normality of oxalic add solution
$= \frac{6.3 \times 1000}{63 \times 250} = 0.4 N$
Now from
$N_{1} V_{1} = N_{2} V_{2}$
$0.4 \times 10 = 0.1 \times V_{2}$
$V_{2} = 40 m L$

Q. An aqueous solution of oxalic acid dihydrate contains its 6.3 g in 250 ml. The volume of 0.1 N NaOH required to completely neutralize 10 ml of this solution

4ml

20ml

2ml

40ml

Normality of oxalic add solution =63×2506.3×1000​=0.4N Now from N1​V1​=N2​V2​ 0.4×10=0.1×V2​ V2​=40mL

Normality of oxalic add solution
$= \frac{6.3 \times 1000}{63 \times 250} = 0.4 N$
Now from
$N_{1} V_{1} = N_{2} V_{2}$
$0.4 \times 10 = 0.1 \times V_{2}$
$V_{2} = 40 m L$