1.

The number of formula units of calcium fluoride CaF_(2) present in 146.4 g of CaF_(2) (The molar mass of CaF_(2) is 78.08 g/mol) is

Answer»

`1.129xx10^(24) CaF_(2)`
`1.146xx10^(24) CaF_(2)`
`7.808xx1-^(24) CaF_(2)`
`1.877xx10^(24) CaF_(2)`

Solution :NUMBER of moles of `CaF_(2)=146.4/78.08=1.875`
`:.` No. of FORMULA UNITS `=1.875xxN_(A)`
`=1.875xx6.023xx10^(23)=11.29xx10^(23)=1.129xx10^(24)`


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