4.9 grams of H2 S o4 is present in hundred ml of the solution then its

1.	4.9 grams of H2 S o4 is present in hundred ml of the solution then its molarity and normality are
Answer» Answer: MOLARITY= 5M NORMALITY=10N Explanation: $molarity = \frac{mass \: of \: solute}{molecular \: mas \: of \: solute \: } \times \frac{1}{volume \: f \: solution \: in \: liters}$ Molarity = $\frac{4.9}{9.8} \times \frac{1000}{100} = 5$ Normality= N×mol with here n is the number of ionizable H ions So, Normality=2× molarity = 2×5 =10N

4.9 grams of H2 S o4 is present in hundred ml of the solution then its molarity and normality are

Answer»

Answer:

Explanation:

$molarity = \frac{mass \: of \: solute}{molecular \: mas \: of \: solute \: } \times \frac{1}{volume \: f \: solution \: in \: liters}$

Molarity =

$\frac{4.9}{9.8} \times \frac{1000}{100} = 5$

Normality= N×mol with

here n is the number of ionizable H ions

So, Normality=2× molarity

= 2×5

=10N