An aqueous solution containing 6.5 g 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

Question

An aqueous solution containing 6.5 g 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 (A)  1000cm3  (B)  2000cm3  (C)  100cm3  (D)  200cm3

Tardigrade Admin · Accepted Answer

Weight of pure NaCl=6.5× 0.9=5.85 text g Number of equivalence of NaCl=(5.85/58.5)=0.1 Number of equivalence of NaOH obtained =0.1 Volume of 1 M acetic acid required for the neutralisation of NaOH=(0.1× 1000/1)=100 text cm3

Q. An aqueous solution containing 6.5 g 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 1M acetic acid required to neutralise $N a O H$ obtained above is

$1000 c m^{3}$

$2000 c m^{3}$

$100 c m^{3}$

$200 c m^{3}$

Weight of pure $N a Cl = 6.5 \times 0.9 = 5.85 g$ Number of equivalence of $N a Cl = \frac{5.85}{58.5} = 0.1$ Number 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 6.5 g 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

1000cm3

2000cm3

100cm3

200cm3