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What does completely ionised mean?

Answer
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Hint:For the salt to be completely ionised, it means that every molecule of the electrolyte must lie in the ionic state.

Complete step-by-step answer:In order to answer the question, let us see the case for dissociation in strong and weak electrolytes.
Weak electrolytes: They have low degrees of ionisation. Hence they have poor ionisation in aqueous solutions which is responsible for their low values of ${{\lambda }_{m}}$ As the dilution increases, the degree of ionisation of weak electrolyte also increases causing more and more ionisation. As a result the value of ${{\lambda }_{m}}$ also increases significantly. When the concentration approaches zero, the degree of ionisation approaches the unit, i.e., the electrolyte almost undergoes complete ionisation. This causes a steep rise in the value of ${{\lambda }_{m}}$ near the infinite dilution.
Strong electrolytes: The strong electrolytes are almost completely ionised in aqueous solution at all concentrations. Hence, the values of their Am are generally high even at high concentrations. However, in concentrated solutions of strong electrolytes the density of ions is quite high which results in the significant inter-ionic interactions. Such inter-ionic attractions effectively reduce the speed of the ions and are responsible for the relatively lower value of ${{\lambda }_{m}}$ at high concentrations. On increasing dilution the ions move apart and interionic attractions are decreased. As a result the value of ${{\lambda }_{m}}$ increases.
Let us consider the following reaction:
\[NaCl+{{H}_{2}}O\to N{{a}^{+}}(aq)+C{{l}^{-}}(aq)\]
For the salt to be completely ionised, it means that every molecule of the electrolyte must lie in the ionic state. After ionisation, the $N{{a}^{+}}$ and $C{{l}^{-}}$ will act as individual particles. For a completely and perfectly ionised solution, we say that the degree of ionisation is equal to 1.

Note:For a completely and perfectly ionised solution, we say that the degree of ionisation is equal to 1.