A) \[6.023\times {{10}^{23}}\]
B) \[6.023\times {{10}^{21}}\]
C) \[6.023\times {{10}^{9}}\]
D) \[6.023\times {{10}^{20}}\]
Correct Answer: D
Solution :
An unknown chlorohydrocarbon has 3.55% of chlorine. 100 g of chlorohydrocarbon has 3.55 g of chlorine. g of chlorohydrocarbon will have \[3.55\times \frac{1}{100}=0.0355\,g\]of chlorine. Atomic wt. of\[Cl=35.5\,g/mol\] Number of moles of \[Cl=\frac{0.0355g}{35.5g/mol}=0.001\,mol\] Number of atoms of \[Cl=0.001\,mol\times 6.023\times {{10}^{23}}mo{{l}^{-1}}=6.023\times {{10}^{20}}\]You need to login to perform this action.
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