0.126 g of acid required 20 mL of 0.1 N NaOH for complete neutralisation. The equivalent mass of an acid is

Question

0.126 g of acid required 20 mL of 0.1 N NaOH for complete neutralisation. The equivalent mass of an acid is (A) 45 (B) 53 (C) 40 (D) 63

Tardigrade Admin · Accepted Answer

Let equivalent mass be E ∴ g-equivalent of acid =(0.126/E) G-equivalents of acid = G equi. of base or (0.126/E)=20×10-3× 0.1 or E=(0.126×103/20×0.1) =63

Q. 0.126 g of acid required 20 mL of 0.1 N NaOH for complete neutralisation. The equivalent mass of an acid is

45

53

40

63

Let equivalent mass be $E$
$∴$ g-equivalent of acid $= \frac{0.126}{E}$
$G$ -equivalents of acid = $G$ equi. of base
or $\frac{0.126}{E} = 20 \times 1 0^{- 3} \times 0.1$
or $E = \frac{0.126 \times 1 0 ^{3}}{20 \times 0.1}$
$= 63$

Q. 0.126 g of acid required 20 mL of 0.1 N NaOH for complete neutralisation. The equivalent mass of an acid is

45

53

40

63

Let equivalent mass be E ∴ g-equivalent of acid =E0.126​ G-equivalents of acid = G equi. of base or E0.126​=20×10−3×0.1 or E=20×0.10.126×103​ =63

Let equivalent mass be $E$
$∴$ g-equivalent of acid $= \frac{0.126}{E}$
$G$ -equivalents of acid = $G$ equi. of base
or $\frac{0.126}{E} = 20 \times 1 0^{- 3} \times 0.1$
or $E = \frac{0.126 \times 1 0 ^{3}}{20 \times 0.1}$
$= 63$