Question

If 49 grams of sulphuric acid is dissolved in 2 litre of solution, then its normality is

Answer 1

g equivalent of mass H2SO4 = 98/2 = 49gm

Volume of solution in liter = 2/1000 = 0.002

no. of g equivalent weight of H2SO4 = 49/49= 1

Normality = no. of g equivalent / volume of

solution in liter.

= 1/0.002

Normality = 500 N

Answer 2

Answer:

The normality of sulphuric acid in the solution is $0.5N$.

Calculation:

Before calculating the normality of the solution, we first have to find the moles and the molarity of sulphuric acid in the solution.

The formula to find out moles:

$Moles=\frac{Mass}{Molar Mass}$

The mass of sulphuric acid- 49g

The molar mass of sulphuric acid- 98g/mol

$Moles = \frac{49}{98} = 0.5mol$ of sulphuric acid

The formula to find molarity:

$Molarity = \frac{Moles}{Volume}$

$Molarity = \frac{0.5}{2} = 0.25 M$ of sulphuric acid.

To calculate the normality, the formula is:

$N = n * M$

Where,

n⇒ n- factor

M⇒ molarity

The n-factor is 2.

$N = 2 * 0.25 = 0.5N$

Conclusion:

When sulphuric acid is dissolved in a solution of 2 liters, its normality will be $0.5N$.