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The number of atoms in `100 g an fcc` crystal with density `d = 10 g//cm^(3)` and the edge equal to 100 pm is equal toA. `1xx10^(25)`B. `2xx10^(25)`C. `3xx10^(25)`D. `4xx10^(25)` |
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Answer» Correct Answer - D `d=(ZxxM_(w))/(a^(3)xxN_(A))` `10=(4xxM_(w))/((10^(-8))xx6xx10^(23))` `6=4xxM_(w)` `(6)/(4)=M_(w)rArr1.5g` `mol = (Mass)(M_(w))` `(100)/(105)` "no. of atoms" `=molxxN_(A)` `(100)/(1.5)xx6xx10^(23)=(6)/(1.5)xx10^(25)=4xx10^(25)` |
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