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A person adds `1*71` gram of sugar `(C_(12)H_(22)O_(11))` in order to sweeten his tea. The number of carbon atoms added are (mol. mass of sugar - 342)A. `3*6xx10^(22)`B. `7*2xx10^(21)`C. `0*05`D. `6*6xx10^(22)` |
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Answer» Correct Answer - A Moles of sugar added `= (1*71)/(342) = 5xx10^(-3)` Carbon atoms added `=12xx5xx10^(-3)xx6*02xx10^(23)` `= 3.61xx10^(22)` |
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