If an iodized salt contains 1 % of KI and a person takes 2 g of the salt every day, the iodide ions going into his body everyday would be approx

Question

If an iodized salt contains 1 % of KI and a person takes 2 g of the salt every day, the iodide ions going into his body everyday would be approx (A)  7.2× 1021  (B)  7.2× 1019  (C)  3.6× 1021  (D)  9.5× 1019

Tardigrade Admin · Accepted Answer

2 g of the salt will contain K I=(2/100) g=0.02 g =(0.02/39+127) mol =1.2 × 10-4 mol K I =1.2 × 10-4 mol I -ions =(1.2 × 10-4)(6.02 × 10-23) I-ions =7.2 × 1019 I-ions

Q. If an iodized salt contains $1%$ of $K I$ and a person takes $2 g$ of the salt every day, the iodide ions going into his body everyday would be approx

$7.2 \times 10^{21}$

$7.2 \times 10^{19}$

$3.6 \times 10^{21}$

$9.5 \times 10^{19}$

$2 g$ of the salt will contain
$K I = \frac{2}{100} g = 0.02 g$
$= \frac{0.02}{39 + 127} m o l$
$= 1.2 \times 1 0^{- 4}$ mol $K I$
$= 1.2 \times 1 0^{- 4}$ mol $I^{-}$ ions
$= (1.2 \times 1 0^{- 4}) (6.02 \times 1 0^{- 23}) I^{-}$ ions
$= 7.2 \times 1 0^{19} I^{-}$ ions

Q. If an iodized salt contains 1% of KI and a person takes 2g of the salt every day, the iodide ions going into his body everyday would be approx

7.2×1021

7.2×1019

3.6×1021

9.5×1019