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Calculate the number of atoms of the constituent elements in 53 g of Na_(2)CO_(3). |
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Answer» SOLUTION :By mole concept, 1 mole of `Na_(2)CO_(3)=" Gram FORMULA mass of "Na_(2)CO_(3)=2xx23+12+16xx3=106g` Now, 106 g of `Na_(2)CO_(3)=" 1 mole"therefore"53 g of "Na_(2)CO_(3)=(1xx53)/(106)=0.5"mole of "Na_(2)CO_(3)` BUT1 mole of `Na_(2)CO_(3)` contains 2 moles of `Na^(+)` ion or `2xx6.022xx10^(23)Na^(+)` ions. `therefore 0.5` mole of `Na_(2)CO_(3)` will CONTAIN `2xx6.022xx10^(23)xx0.5Na^(+)" ions"=6.022xx10^(23)Na^(+)ions` Again, 1 mole of `Na_(2)CO_(3)` contains 1 mole of carbon atoms `=6.022xx10^(23)Na^(+)` ions `therefore 0.5`, mole of `Na_(2)CO_(3)` will contain `=6.022xx10^(23)xx0.5` carbon atoms `=3.011xx10^(23)" carbon atoms"` Further, 1 mole of `Na_(2)CO_(3)` contains 3 moles of oxygen atoms of `3xx6.022xx10^(23)` oxygen atoms `therefore 0.5` mole of `Na_(2)CO_(3)` will contain `=3xx6.022xx10^(23)xx0.5` oxygen atoms `=9.033xx10^(23)` oxygen atoms |
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