Question

Calculate factor for each ml of 0.1 N Iodine Solution by the aid of Sodium Thiosulphate (248.2 g)

Answer 1

....

$\tt{\huge{\purple{ ....................... }}}$

CORRECT QUESTION -

Calculate the n - factor for each ml of 0.1 N Iodine Solution by the aid of Sodium Thiosulphate (248.2 g).

CONCEPT USED -

Mole Concept , N - factor

SOLUTION -

From the above Question, We can get the following information.....

Here, the given Iodine Solution is 0.1 Normal .

The amount of Sodium Thiosulfate is 248.2 gram .

According to the law of equivalents ,

Number Of Equivalents of Reactant = Number Of Equivalents Of Product .

Number Of Equivalents = Number Of Moles × N factor.

Let us first find the number Of Moles Of Sodium Thiosulfate.

Given Mass of Sodium Thiosulfate = 248.2 grams

Molar Mass Of Sodium Thiosulfate = 158.11 grams / mol

Now :

Number Of Moles = [ Given Mass ] / [ Molar Mass ]

=> [ 248.2 ] / [ 158.11 ]

=> Which is approximately equal to 1.57 moles .

Now, the N factor Of Sodium Thiosulfate in both acidic and basic mediums is 2.

So,

Number Of Equivalents Of Sodium Thiosulfate

=> 1.57 × 2 = 3.14

Number Of Equivalents is also equal to :

=> Normality × Volume × N factor

Number Of Equivalents Of Iodine Solution :

=> 1 × 1 × 10 ^ (-3) × N Factor

{ As We have to calculate for each ml }

=> N factor Of Iodine Solution => 3.14 × 10 ^ 3 = 3140.

[ This is not possible practically. The values given in the Question is Wrong. ]

ANSWER -

The N - factor for each ml of 0.1 N Iodine Solution is 3140.

______________________________________________________

ADDITIONAL FORMULAE -

• Equivalents Of Reactants = Equivalents Of Products

• Number Of Equivalents :

• => Number Of Moles × N Factor

• => Normality × Volume × N Factor

• => Molarity × N factor × Volume

• Number Of Moles :

• => [ Given Mass ] / [ Molar Mass ]

______________________________________________________

SIMILAR QUESTIONS -

• https://brainly.in/question/1747569

• https://brainly.in/question/3191532

Answer 2

Question :-

Calculate n-factor for each ml of 0.1 N Iodine Solution by the aid of Sodium Thiosulphate (248.2 g).

Solution :-

The given Iodine Solution is 0.1 Normal .

The amount of Sodium Thiosulphate is 248.2 gram.

By using the Law of Equivalents,

⇒  Number Of Equivalents of Reactants = Number Of Equivalents Of Products

⇒Number Of Equivalents = Number of Moles × N factor

Now,

we know, Molar Mass Of Sodium Thiosulfate = 158.11 grams / mol

Number Of Moles = [ Given Mass ] / [ Molar Mass ]

⇒ Number Of Moles =  [ 248.2 ] / [ 158.11 ]

⇒ Number Of Moles =  1.57 (approx.)

Number Of Equivalents Of Sodium Thiosulphate  = 1.57*2 = 3.14

Number Of Equivalents is also equal to :

Normality × Volume × N factor

Number Of Equivalents Of Iodine Solution :

= 1 × 1 × 10⁻³ × N Factor

(As We have to calculate for each ml)

⇒ N factor Of Iodine Solution => 3.14 × 10³ = 3140

∴ The N - factor for each ml of 0.1 N Iodine Solution is 3140.