Chapter 15.3, Problem 7RQ

Interpretation Introduction

Interpretation: Whether $1M{H}_{2}S{O}_4$ is same as $1N{H}_{2}S{O}_4$ or not needs to be determined.

Concept introduction: Molarity can be defined as the number of moles of solute in 1 litre of solution. Thus,

  $\text{Molarity(}M\text{)}=\frac{\text{Number of moles of solute}}{\text{Volume of the solution in litre}}$

Normality mainly used in acid-base reaction where it gives more in-depth information of the concentration of the solution.

  $\text{Normality}\left(N\right)=\frac{\text{grams equivalent of substance}}{\text{volume of the solution in litre}}$

Molecules reactive capacity is known as $\text{grams equivalent}$ weight.

Simple relation between normality and molarity is as follows,

  $\text{Normality}\left(N\right)=n\text{ (number of equivalent)}\times M\left(\text{Molarity}\right)$

Answer to Problem 7RQ

  $1M{H}_{2}S{O}_4$ is not same as $1N{H}_{2}S{O}_4$ .

Explanation of Solution

1 mol of $H_{2}S{O}_4$ dissociates to give 2 moles of $H^{+}$ and hence $H_{2}S{O}_4$ is diacidic.

Normality of $H_{2}S{O}_4$

  $=\text{acidity }\times \text{ molarity}$

  $=(2\times \text{1) }N$

  $=2N$

Conclusion

  $1M{H}_{2}S{O}_4$ is not same as $1N{H}_{2}S{O}_4$ .