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What is the total number of atoms present in 25.0 mg of camphor, `C_(10)H_(16)O`?A. `9.89xx10^(19)`B. `6.02xx10^(20)`C. `9.89xx10^(20)`D. `2.67xx10^(21)` |
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Answer» Correct Answer - D Molar mass of `C_(10)H_(16)O = 12+16+16 = 152 g mol^(-1)` `25.0 mg = 25.0 mgxx((1)/(10^(3)mg))xx((1mol)/(152g))xx(27xx6.02xx10^(23)atoms)/(1mol)` `=(25xx27xx6.02)/(152)xx10^(20)` atom `= 26.73xx10^(20)` atom `= 2.67xx10^(21)` atoms |
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