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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`B. `6.022xx10^23`C. `1xx10^23`D. `12.044xx10^23` |
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Answer» Molarity `=("No. of moles of solute" xx 1000)/("Volume of solution (in mL)")` `therefore " 1M " H_2SO_4` contains 1 mole of `H_2SO_4` in 1000 mL of solution `therefore " 0.02M " H_2SO_4` in 100 mL of solution contains `(0.02xx100)/(1000)=2xx10^(-3)` mol `=2xx10^(-3) xx 6.022xx10^23=12.044xx10^20` molecules |
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