In a titration, 25 cm3 of 0.1 N oxalic acid solution requires 20 cm3 of sodium hydroxide for complete neutralization. The mass of sodium hydroxide in 250 cm3 solution is

Question

In a titration, 25 cm3 of 0.1 N oxalic acid solution requires 20 cm3 of sodium hydroxide for complete neutralization. The mass of sodium hydroxide in 250 cm3 solution is (A) 12.5 g (B) 1.25 g (C) 0.125 g (D) 125 g

Tardigrade Admin · Accepted Answer

Oxalic acid                                          NaOH V1 = 25 cm3, N1 = 0.1 N     V2 = 20 cm3, N2 =? N1V1 = N2V2 0.1 × 25 = N2 × 20 N2 = 0.125 N N = (w × 100/Eq.wt × V) ⇒ 0.125 = (w × 1000/40 × 250) ⇒ : : w = 1.25 g

Q. In a titration, 25 $c m^{3}$ of 0.1 N oxalic acid solution requires 20 $c m^{3}$ of sodium hydroxide for complete neutralization. The mass of sodium hydroxide in 250 $c m^{3}$ solution is

12.5 g

1.25 g

0.125 g

125 g

Oxalic acid NaOH
$V_{1} = 25 c m^{3}, N_{1} = 0.1 N$ $V_{2} = 20 c m^{3}, N_{2} = ?$
$N_{1} V_{1} = N_{2} V_{2}$
$0.1 \times 25 = N_{2} \times 20$
$N_{2} = 0.125 N$
$N = \frac{w \times 100}{Eq . wt \times V} \Rightarrow 0.125 = \frac{w \times 1000}{40 \times 250}$
$\Rightarrow w = 1.25 g$

Q. In a titration, 25 cm3 of 0.1 N oxalic acid solution requires 20 cm3 of sodium hydroxide for complete neutralization. The mass of sodium hydroxide in 250 cm3 solution is