Question

6. 10 mL of 10 M H2SO4 is diluted to 250 mL, The strength of the diluted solution is (a) 0.80 N (b) 0.40 N (c) 1.0N (d) 0.60 N​

Answer 1

Answer :-

The strength of the diluted solution is 0.80 N [Option.a]

Explanation :-

Here, the initial volume (V₁) of H₂SO₄ was 10 mL which is diluted to 250 mL (V₂) . Also, the molarity of the solution before dilution was 10 M.

So firstly, let's calcualte the initial normality of the solution. (N factor of H₂SO₄ is 2) .

$\boxed{\bf{ \textbf{Normality} = Molarity \times N \: \: factor}}$

$\implies N_1 = 10 \times 2\\\\\implies N_1 = 20 \: \: N$

We have got the initial normality of the solutiom. So now, we can calculate the final normality (N₂) by using the formula of normality based on volumetric analysis.

$\implies N_1V_1 = N_2V_2\\\\\implies 20 \times 10 = N_2 \times 250\\\\\implies 200 = 250N_2\\\\\implies N_2 = \dfrac{200}{250}\\\\\:\implies N_2 = 0.80 \: \: N$

Answer 2

Given :

Intial Volume of H₂SO₄ (V₁) = 10ml

10 ml is diluted into (V₂) = 250 ml

Molarity of the Solution before dilution (M) = 10M

To Find :

Final Normality of the Solution i.e., after dilution (N₂)

Solution :

Calculating Intial Normality of the Solution :

By using, Normality = Molarity * N-Factor

⇒ N₁ = 10 * 2

⇒ N₁ = 20N

Calculating Final Normality of the Solution :

By using,

N₁V₁ = N₂V₂

⇒ 20 * 10 = N₂ * 250

⇒ N₂ = 200/250

⇒ N₂ = 0.8N

$\:$

FINAL ANSWER :

The Strength of the Diluted Solution is 0.80N .

Option (a) is the correct answer.