3.5 g of a mixture of NaOH and KOH were dissolved and made up to 250 mL .25 mL of this solution were completely neutralised by 17 mL of ( N / 2) HCl solution. Then, the percentage of KOH in mixture is

Question

3.5 g of a mixture of NaOH and KOH were dissolved and made up to 250 mL .25 mL of this solution were completely neutralised by 17 mL of ( N / 2) HCl solution. Then, the percentage of KOH in mixture is (A) 80 (B) 10 (C) 34 (D) 56

Tardigrade Admin · Accepted Answer

Suppose the amount of NaOH in the mixture =x g ∴ Amount of KOH =(3.5-x) g 25 mL of alkali solution ≡ 17 mL of ( N / 2) HCl or, 250 mL of alkali solution ≡ 170 mL of ( N / 2) HCl 1000 mL of ( N ) HCl contain 36.5 g of HCl 170 mL of N / 2 HCl will contain (36.5 × 170/1000 × 2)=3 ⋅ 1025 g of HCl Eq. mass of NaOH =40 and that of KOH =56 40 g of NaOH require 36.5 g of HCl x g of NaOH will require (36.5 × x/40) g of HCl Similarly, 56 g of KOH require 36.5 g of HCl (3.5-x) g of KOH will require (36.5(3.5-x)/56) g of HCl Thus, (36.5 x/40)+(36.5(3.5-x)/56) =3.1025 g of HCl Or x=3.15 g = Amount of NaOH in the mixture ∴ Amount of KOH =(3 ⋅ 5-3 ⋅ 15) g =0 ⋅ 35 g 3.5 g of the mixture contain 0.35 g of KOH 100 g of the mixture contain =(0 ⋅ 35 g × 100/35) =10 g of KOH Hence, % of KOH =10

Q. 3.5g of a mixture of NaOH and KOH were dissolved and made up to 250mL.25mL of this solution were completely neutralised by 17mL of (N/2)HCl solution. Then, the percentage of KOH in mixture is

80

10

34

56

Q. $3.5 g$ of a mixture of $N a O H$ and $K O H$ were dissolved and made up to $250 m L .25 m L$ of this solution were completely neutralised by $17 m L$ of $(N /2) H Cl$ solution. Then, the percentage of $K O H$ in mixture is