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Alcohol level in blood is determined by the reaction with K_2Cr_2O_7 solution in acidic medium. Calculate the blood level in mass percent if 10 " mL of " 0.05 M solution of K_2Cr_2O_7 is required for the reaction of a 10.0 g sample of blood.

Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Mole method: <br/> Balance the redox reaction of `C_2H_5OH` with `Cr_2O_7^(2-)` in acidic <a href="https://interviewquestions.tuteehub.com/tag/medium-1092763" style="font-weight:bold;" target="_blank" title="Click to know more about MEDIUM">MEDIUM</a> (Mw of `C_2H_5OH=46g)` <br/> `cancel6e^(-)+Cr_2O_7^(2-)to2Cr^(3+)]xx2` <br/> `underline(C_2H_5OHto2CO_2+cancel(12e^(-))` <br/> `underline(C_2H_5OH+2Cr_2O_7^(2-)to4Cr^(3+)+2CO_2)` <br/> For 2 " mol of "`Cr_2O_7^(2-)-=<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a> " mol of "C_2H_5OH` is required `10xx0.05xx10^(-3) " mol of "Cr_2O_7^(2-)` <br/> `-=(10xx0.05xx10^(-3))/(2) " mol of "C_2H_5OH` <br/> `-=25xx10^(-5)` " mol of "`C_2H_5OH` <br/> `-=25xx10^(-5)xx46g of C_2H_5OH` <br/> Percent of alcohol `-=(25xx10^(-5)xx46)/(10)xx100=0.115%` <br/> Equivalent method: <br/> `Cr_2O_7^(2-)-=C_2H_5OH` <br/> `1mEq-=1mEq` <br/> `10xx0.05xx6m" Eq of "Cr_2O_7^(2-)=3 mEq-=3 m" Eq of "C_2H_5OH` <br/> `3 mEq-=3 mEq` <br/> `3m" Eq of "C_2H_5OH-=3xx10^(-3)eq-=3xx10^(-3)xx(<a href="https://interviewquestions.tuteehub.com/tag/46-317673" style="font-weight:bold;" target="_blank" title="Click to know more about 46">46</a>)/(12)g` <br/> Percent of alcohol `-=(3xx10^(-3)xx46xx100)/(<a href="https://interviewquestions.tuteehub.com/tag/12xx10-1784290" style="font-weight:bold;" target="_blank" title="Click to know more about 12XX10">12XX10</a>)-=0.115%`</body></html>


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