Conc. $ {{\mathbf{H}}_2}{\mathbf{S}}{{\mathbf{O}}_4} $ can not be used to prepare pure $ {\mathbf{HBr}} $ from $ {\mathbf{NaBr}} $ as it reacts slowly with $ {\mathbf{NaBr}} $ .
(A) True
(B) False

Hint: To answer the given question, we first need to understand what is reducing agent and oxidizing agent. Also how a strong reducing agent and a strong oxidizing agent affects the reactivity of the reaction. If an oxidizing or reducing agent used is weak or strong, then the product we get is, according to the reactivity of the reducing or oxidizing agent

Complete step by step answer:
Redox is a very important kind of reaction in chemistry. It is a short form for reduction and oxidation reaction. Always remember when there is a reduction simultaneously there occurs or an oxidation reaction.
 In simple words, the reduction is understood as the addition of hydrogen or an electropositive element to that compound, according to electrons, the reduction occurs when a compound gains electrons. On the other hand, oxidation is understood as the addition of oxygen or any electronegative element to that compound, according to electrons, oxidation occurs when a compound loses electrons.
Now, in the reaction of concentrated sulphuric acid and sodium bromide, hydrogen bromide is a strong reducing agent. Therefore, hydrogen bromide will reduce the concentrated sulphuric acid to sulphur dioxide, and will itself oxidise to form bromine gas.
 $ NaBr + {H_2}S{O_4} \to NaHS{O_4} + HBr $
 $ {H_2}S{O_4} + 2HBr \to S{O_2} + B{r_2} + 2{H_2}O $
And hence concentrated sulphuric acid does not prepare hydrogen bromide with sodium bromide because hydrogen bromide is a strong reducing agent and not because the reaction is slow.
Correct option is B; false.

Note:
The oxidation and reduction of the reaction can also be calculated by the oxidation number. The oxidation number is written normally above the symbol of the atom. If the oxidation number decreases it means it is the reduction and if there is an increase in the oxidation number, it means there is an oxidation of that atom.