Solveeit Logo
Login

Question

Question: If an iodized salt contains \[1\% \] of KI and a person takes \(2\)g of the salt every day, the iodi...

If an iodized salt contains 1%1\% of KI and a person takes 22g of the salt every day, the iodide ions going into his body everyday would be appropriate-
A.7.2×10217.2 \times {10^{21}}
B.7.2×10197.2 \times {10^{19}}
C.3.6×10213.6 \times {10^{21}}
D.9.5×10199.5 \times {10^{19}}

Explanation

Solution

First, we will find the mass of KI in the iodized salt. Then we will find the moles of Iodide ion in the salt using the formula-
Moles of Iodide ion=Given mass of KIMolecular mass of KI\dfrac{{{\text{Given mass of KI}}}}{{{\text{Molecular mass of KI}}}}. We know that molecular mass of KI=molecular mass of iodine and molecular mass of potassium. Find the molecular mass then put the given values in the formula and solve it. Then to find the number of iodide ions, multiply the moles by Avogadro's number.

Complete step by step answer:
Given, an iodized salt contains 1%1\% of KI
The amount of salt a person takes everyday=22g
We have to find the iodide ions going into his body every day.
First we will find the mass of KI in the salt so 22g of salt will contains KI=1%1\% of 2g2g
On solving, we get-
The mass of KI in salt=2100=0.02\dfrac{2}{{100}} = 0.02 g
Now we know that KI contains one potassium ion and one iodide ion.
So we can write the moles of Iodide ion=Given mass of KIMolecular mass of KI\dfrac{{{\text{Given mass of KI}}}}{{{\text{Molecular mass of KI}}}}
On putting the values we get-
Moles of Iodide ion=0.0239+127\dfrac{{0.02}}{{39 + 127}}
On solving, we get-
Moles of Iodide ion=0.02166\dfrac{{0.02}}{{166}}
On further solving, we get-
Moles of Iodide ion= 1.2×10−41.2 \times {10^{ - 4}}
Now, we know that the number of Iodide ions= Moles of iodide ion × Avogadro number-- (i)
And we know that Avogadro number=6.023×10236.023 \times {10^{23}}
On putting the given values in eq. (i), we get-
Number of Iodide ions= 1.2×10−4×6.023×10231.2 \times {10^{ - 4}} \times 6.023 \times {10^{23}}
Now we know that mn×ma=mn+a{m^n} \times {m^a} = {m^{n + a}} then on applying this we get-
Number of Iodide ions= 1.2×6.023×10−4+231.2 \times 6.023 \times {10^{ - 4}}^{ + 23}
On solving, we get-
Number of Iodide ions= 7.2×10197.2 \times {10^{19}}

The correct answer is option B.

Note:
Here the student may get confused as to why we used the formula-
Moles of Iodide ion=Given mass of KIMolecular mass of KI\dfrac{{{\text{Given mass of KI}}}}{{{\text{Molecular mass of KI}}}}
So remember when one KI breaks into its ion it produces one mole of Iodide ion and one mole of Potassium ion. This means one mole of KI will produce the one mole of Iodide ion and we know the formula of moles is the given mass divided by molecular mass.
Then the number of moles of Iodide ion will be equal to the number of moles of potassium iodide.