Question

5.
A 5 M solution of H, SO, is diluted from 1 litre to a volume of 100 litres, the normality of the solution will be
(4) 0.02 mol
1 N
(2) SN
(3) 0.1 N
(4) 0.5 N
(1) 1N
but how nfactor was 2​

Answer 1

QUESTION : A 5 M solution of H2SO4 is diluted from 1 litre to a volume of 100 litres, the normality of the solution will be

Given :

• Initial molarity = 5M
• Initial volume = 1L
• Final volume = 100L

To find :

• Normality of final solution

Important point & formula :

• M{1} × V{1} = M{2} × V{2}
• N{1} × V{1} = N{2} × V{2}
• Normality = Molarity × n factor

here,

M{1} = Initial molarity

V{1} = Initial Volume

M{2} = Final Molarity

V{2} = Final volume

N{1} = Initial normality

N{2} = Final normality

Solution :

$\sf \: H_{2}SO_{4}\: <=> \: 2H^{+} \: +\:SO_{4}^{ - 2}$

therefore , n factor for acid = number of positive charge

here n factor = 2

{as there is +2 charge in H2SO4}

We can solve this question in 2 way

1. Find final molarity and then normality through it
2. First find initial normality from molarity then find final molarity

1st method ) using , M{1} × V{1} = M{2} × V{2}

putting respective values we get;

=> 5×1 = M{2} × 100

=> M{2} = 5/100 = 0.05M

now we know Normality = Molarity×nfactor

Final Normality = Final Molarity × nfactor

Final Normality = 0.05 ×2

Final Normality = 0.1 N

2nd Method )

We know Normality = Molarity×nfactor

Initial Normality = initial Molarity × nfactor

Initial Normality = 5 ×2

Initial Normality = 10 N

Also , N{1} × V{1} = N{2} × V{2}

putting respective values, we get;

=> 10 × 1 = N{2} × 100

=> N{2} = 10/100

=> N{2} = 0.1N

ANSWER : Option (3) 0.1N