The number of atoms in 4.25 g of `NH_(3)` is approximatelyA. `1xx10^(23)`B. `1.5xx10^(23)`C. `2xx10^(23)`D. `6xx10^(23)`

The number of atoms in 4.25 g of `NH_(3)` is approximatelyA. `1xx10^(2

1.	The number of atoms in 4.25 g of `NH_(3)` is approximatelyA. `1xx10^(23)`B. `1.5xx10^(23)`C. `2xx10^(23)`D. `6xx10^(23)`
Answer» Correct Answer - D Total no. of atoms in 4.25g of `NH_(3)` `= (4xx6.02xx10^(23))/(17)xx4.25 = 6xx10^(23)`

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The number of atoms in 4.25 g of `NH_(3)` is approximatelyA. `1xx10^(23)`B. `1.5xx10^(23)`C. `2xx10^(23)`D. `6xx10^(23)`

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