How much of NaOH is required to neutralize 1500 cm3 of 0.1 N HCl?

Question

How much of NaOH is required to neutralize 1500 cm3 of 0.1 N HCl? (A) 60 g (B) 6 g (C) 4 g (D) 40 g

Tardigrade Admin · Accepted Answer

1500 cm3 of 0.1 N HCl ≡ 1500 cm3 of 0.1 N NaOH ≡ 1500 cm3 of 0.1 M NaOH = ( (1500/1000) L ) × (0.1 mol L-1 ) = 0.15 mol. = 0.15 mol × 40 g mol-1 = 6 g

Q. How much of NaOH is required to neutralize 1500 $c m^{3}$ of 0.1 N HCl?

60 g

6 g

4 g

40 g

1500 $c m^{3}$ of 0.1 N HCl $\equiv$ 1500 $c m^{3}$ of 0.1 N NaOH
$\equiv$ 1500 $c m^{3}$ of 0.1 M NaOH
= $(\frac{1500}{1000} L) \times (0.1 m o l L^{- 1})$ = 0.15 mol.
= 0.15 mol $\times$ 40 g mol $^{- 1}$ = 6 g

Q. How much of NaOH is required to neutralize 1500 cm3 of 0.1 N HCl?

60 g

6 g

4 g

40 g

1500 cm3 of 0.1 N HCl ≡ 1500 cm3 of 0.1 N NaOH ≡ 1500 cm3 of 0.1 M NaOH = (10001500​L)×(0.1molL−1) = 0.15 mol. = 0.15 mol × 40 g mol−1 = 6 g

Q. How much of NaOH is required to neutralize 1500 $c m^{3}$ of 0.1 N HCl?

1500 $c m^{3}$ of 0.1 N HCl $\equiv$ 1500 $c m^{3}$ of 0.1 N NaOH
$\equiv$ 1500 $c m^{3}$ of 0.1 M NaOH
= $(\frac{1500}{1000} L) \times (0.1 m o l L^{- 1})$ = 0.15 mol.
= 0.15 mol $\times$ 40 g mol $^{- 1}$ = 6 g