A) \[7.2\times {{10}^{21}}\]
B) \[7.2\times {{10}^{19}}\]
C) \[3.6\times {{10}^{21}}\]
D) \[9.5\times {{10}^{19}}\]
Correct Answer: B
Solution :
2g of the salt will contain\[KI=\frac{2}{100}g=0.02g\] \[=\frac{0.02}{39+127}\]mol\[=1.2\times {{10}^{-4}}mol\text{ }KI\] \[=1.2\times {{10}^{-4}}mol\text{ }{{\text{I}}^{-}}\text{ }ions\] \[=(1.2\times {{10}^{-4}})(6.02\times {{10}^{-23}}){{I}^{-}}\] ions \[=7.2\times {{10}^{19}}{{I}^{-}}ions\]
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