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The number of `H_(2)O` molecules in a drop of water weighing 0.018 g is :A. `6.022 xx 10^(26)`B. `6.022 xx 10^(23)`C. `6.022 xx 10^(19)`D. `6.022 xx 10^(20)` |
Answer» Correct Answer - D Molar mass of water `(H_(2)O)=18 g` No. of `H_(2)O` drops present in 18g of water `= 6.022 xx 10^(23)` No. of `H_(2)O` drops present in 0.018 of water `= ((0.018g))/((18.0g))xx6.022xx10^(23)=6.022xx10^(20)`. |
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