Question

The total no of ions present in 1ml of 0.1M barium nitrate \[Ba{\left( {N{O_3}} \right)_2}\] solution is:
\[
  A.\;6.02 \times {10^{18}} \\
  B.\;6.02 \times {10^{19}} \\
  C.\;3 \times 6.02 \times {10^{19}} \\
  D.\;3 \times 6.02 \times {10^{18}} \\
\]

Answer 1

Hint: Whenever we are asked to calculate the number of ions, first we need to calculate the number of moles present in a given element. If you calculate the number of moles then the number of ions will be the product of no of moles and Avogadro's number. Number of ions = \[n \times {N_A}\]

Complete answer:
Now, we came to know that 1 mole of any substance contains Avogadro's number of particles. In the question, the no of moles are not given, instead 1ml of 0.1M solution is given. So by using these values, we need to find the number of moles present. The formula is given by
Number of moles = Molarity x Volume (L)
Given, molarity=0.1M and volume = 1ml (0.001L)
Substituting the values, we get
No. of moles = 0.1 x 0.001
\[ = {10^{ - 4}}\]
When Barium nitrate undergoes ionization, it gives three ions. The reaction is as follows
\[Ba{\left( {N{O_3}} \right)_2} \to B{a^{2 + }} + N{O_3}^{2 - }\]
So here, in one mole of barium nitrate, three ions are present and one ion contains Avogadro no of particles, so 3 ions contain \[3 \times 6.023 \times {10^{23}}\]. Then how many ions will be present in \[{10^{ - 4}}\] moles?
1mole \[ \to 3 \times 6.023 \times {10^{23}}\]
\[{10^{ - 4}} \to \] ?
\[
   = 3 \times 6.023 \times {10^{23}} \times {10^{ - 4}} \\
   = 3 \times 6.023 \times {10^{19}} \\
\]
And hence the correct option is C.

Note:
The scientist named Amedeo Avogadro has introduced a proportionality constant called Avogadro's constant. It is denoted by \[{N_A}\] . Avogadro’s constant gives the number of particles present in one mole of a given substance which is equal to \[6.023 \times {10^{23}}\] . Here, the substance may be an atom or an ion or any molecule.