Question

If 9.8g of h2so4 are present in 500 ml find normality

Answer 1

Normality=no. of gram equivalents /vol of solution in L

=mass×1000/equivalent wt × vol of sol(ml)

=9.8×1000/49×500

= 0.4 eq/l

Answer 2

The normality of sulfuric acid is 0.4 grams

Explanation:

Normality is defined as the umber of gram equivalents dissolved per liter of the solution.

Mathematically,

$\text{Normality of solution}=\frac{\text{Given mass}\times 1000}{\text{Equivalent mass}\times \text{Volume of solution (in mL)}}$

We are given:

Given mass of sulfuric acid = 9.8 g

Equivalent mass of sulfuric acid = $\frac{\text{Molecular mass}}{\text{Basicity}}=\frac{98}{2}=49g/eq$

Volume of solution = 500 mL

Putting values in above equation, we get:

$\text{Normality of solution}=\frac{9.8\times 1000}{49g/eq\times 500mL}\\\\\text{Normality of solution}=0.4N$

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