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No. of `H_(2)O` molecules in a drop of water weighing 0.05 g is :A. `1.15 xx 10^(23)`B. `1.672 xx 10^(21)`C. `1.5 xx 10^(20)`D. `6.022 xx 10^(22)` |
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Answer» Correct Answer - B 18.0 g of `H_(2)O` contain molecules `=6.022 xx 10^(23)` 0.05 g of `H_(2)O` contain molecules `= ((0.05g))/((18.0g))xx6.022xx10^(23)` `=1.673xx10^(21)` molecules. |
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