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The degree of dissociation of water at 25oC is 1.8×107% and density is 1.0gcm3. The ionic constant for water is:
A.1.0×1014
B.2.0×1016
C.1.0×1016
D.1.0×108

Answer
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Hint: Concentration of water is the ratio of its moles per litre. Ionization constant is the product of degree of dissociation and concentration of water. It is represented by Ka. We shall calculate the concentration of 1 litre of water from the density and substitute it in the ionization constant formula.
Formula Used: Ka=Cα2
Where C is the concentration of water, α is the degree of dissociation and Ka is the ionization constant.

Complete step by step answer:
Water is a weak electrolyte and thus, cannot ionize completely.
First we need to calculate the concentration of water which is known to be the moles per litre. The density of water is given as 1.0gcm3.
Thus, the mass of one litre of water which is 1000cm - 3 is equal to 1000g. The molar mass of water is known as 18g.
We know that the concentration of water is given as 100018=55.5molL - 1
Water dissociates and gives hydrogen ion and hydroxyl anion. The ionic product is(Kw)=[H+][OH]=C2α2
The ionic constant is given by Ka=[H+][OH][H2O]=C2α2C
Thus we get, Cα2=(55.56)×(1.8×109)22×1016

Therefore, the correct answer is B.

Note:
When the product of hydrogen ion concentration and hydroxide ion concentration is constant when temperature is kept constant is called as the ionic product of water. Water undergoes self- ionization forming hydronium ions and hydroxide ions. When two molecules of water collide in the liquid state, hydrogen ions are transferred from one molecule to the other. While writing the expression of an ionic product, concentration of water is not included because it is a pure liquid and its concentration can be considered equal to unity. At room temperature the ionic product of water is generally equal to 1014.The unit of ionic product is mol2L - 2.