An aqueous solution containing 6.5 gm of NaCl of 90 % purity was subjected to electrolysis. After the complete electrolysis, the solution was evaporated to get solid NaOH. The volume of 1 M acetic acid required to neutralise NaOH obtained above is

Question

An aqueous solution containing 6.5 gm of NaCl of 90 % purity was subjected to electrolysis. After the complete electrolysis, the solution was evaporated to get solid NaOH. The volume of 1 M acetic acid required to neutralise NaOH obtained above is (A) 100 cm3 (B) 200 cm3 (C) 1000 cm3 (D) 2000 cm3

Tardigrade Admin · Accepted Answer

Weight of pure NaCl =6.5 × 0.9=5.85 g No. of equivalence of NaCl =(5.85/58.5)=0.1 No. of equivalence of NaOH obtained =0.1 Volume of 1 M acetic acid required for the neutralisation of NaOH =(0.1 × 1000/1) =100 cm 3

Q. An aqueous solution containing $6.5 g m$ of $N a Cl$ of $90%$ purity was subjected to electrolysis. After the complete electrolysis, the solution was evaporated to get solid $N a O H$ . The volume of $1 M$ acetic acid required to neutralise $N a O H$ obtained above is

$100 c m^{3}$

$200 c m^{3}$

$1000 c m^{3}$

$2000 c m^{3}$

Weight of pure $N a Cl = 6.5 \times 0.9 = 5.85 g$

No. of equivalence of $N a Cl = \frac{5.85}{58.5} = 0.1$

No. of equivalence of $N a O H$ obtained $= 0.1$

Volume of $1 M$ acetic acid required for the

neutralisation of $N a O H = \frac{0.1 \times 1000}{1}$

$= 100 c m^{3}$

Q. An aqueous solution containing 6.5gm of NaCl of 90% purity was subjected to electrolysis. After the complete electrolysis, the solution was evaporated to get solid NaOH. The volume of 1M acetic acid required to neutralise NaOH obtained above is

100cm3

200cm3

1000cm3

2000cm3