Question

If an iodised salt contains $1$% of $KI$ and a person takes $2g$ of the salt every day, the iodide ions going into his body every day would be approximate:

• A. $7.2\times {10}^{21}$
• B. $7.2\times {10}^{19}$
• C. $3.6\times {10}^{21}$
• D. $9.5\times {10}^{19}$

Answer 1

The correct option is B $7.2\times {10}^{19}$

$2g$ of the salt will contain $KI=\frac{1}{100}\times 2g=0.02g$

Moles of $I^{-}=\frac{0.02}{39+127}mol$

$=1.2\times {10}^{-4}molKI$

$=(1.2\times {10}^{-4})mol{I}^{-}ions$

Number of Iodide ions $=(1.2\times {10}^{-4})(6.02\times {10}^{23})I^{-}ions$

$=7.2\times {10}^{19}{I}^{-}ions$.

Hence, option B is correct.