Question

0.5M h2so4 is diluted from 1litre to 10 litre,normality of the resulting solution is​

Answer 1

M1 = 0.5

V1 = 1L

n factor = 2

N1 = M1 X n factor = 0.5 X 2 = 1

V2 = 10L

The moles will be same in both cases,

N1 X V1 = N2 X V2

1 x 1 = N2 X 10

N2 = 1/10 = 0.1

Answer 2

The normality of the resulting solution is 0.1 N

Explanation:

We know that:

$\text{Normality}=n\times \text{Molarity}$

where,

n = basicity for acids = 2 (for sulfuric acid)

Molarity of sulfuric acid = 0.5 M

Putting values in above equation:

$\text{Normaity}=2\times 0.5=1.0N$

To calculate the normality of the diluted solution, we use the equation:

$N_1V_1=N_2V_2$

where,

$N_1\text{ and }V_1$ are the normality and volume of the concentrated sulfuric acid solution

$N_2\text{ and }V_2$ are the normality and volume of diluted sulfuric acid solution

We are given:

$N_1=1.0N\\V_1=1L\\N_2=?N\\V_2=10L$

Putting values in above equation, we get:

$1.0\times 1=N_2\times 10\\\\N_2=\frac{1.0\times 1}{10}=0.1N$

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