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One mole of any substance contains `6.022xx10^(23)` atoms/molecules. Number of molecules of `H_(2)SO_(4)` present in 100mL of 0.02M `H_2SO_(4)` solution isA. `12.044xx10^(20)` moleculesB. `6.022xx10^(23)` molecuelsC. `1xx10^(23)` molecuelsD. `12.044xx10^(23)` molecules |
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Answer» One mole of any substance contains `6.022xx10^(23)` atoms/molecuels. Hence, Number of millimoles of `H_(2)SO_(4)` =molarity xx volume in mL `=0.02xx100=2` millimoles `=2xx10^(-3)` mol Number of molecuels =number of moles `xxN_(A)` `=2xx10^(-3)xx6.022xx10^(23)` `=12.044x10^(20)` molecuels |
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