A) \[1.129\times {{10}^{24}}Ca{{F}_{2}}\]
B) \[1.146\times {{10}^{24}}Ca{{F}_{2}}\]
C) \[7.808\times {{10}^{24}}Ca{{F}_{2}}\]
D) \[1.877\times {{10}^{24}}Ca{{F}_{2}}\]
Correct Answer: A
Solution :
\[Ca{{F}_{2}}=146.4g\] Molecular weight of \[Ca{{F}_{2}}=78.08g/mol\] Moles of \[Ca{{F}_{2}}=\frac{wt}{mo.\,wt}\] \[=\frac{146.4}{78.08}=1.875mol\] Number of \[Ca{{F}_{2}}\] atoms in 146.4 g of \[Ca{{F}_{2}}\]= No. of moles \[\times 6.022\times {{10}^{23}}\] \[=1.875\times 6.022\times {{10}^{23}}\] \[=11.29\times {{10}^{23}}\] \[=11.29\times {{10}^{24}}Ca{{F}_{2}}\]
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