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.

Explanation

Mole method:
Balance the redox reaction of `C_2H_5OH` with `Cr_2O_7^(2-)` in acidic medium (Mw of `C_2H_5OH=46g)`
`cancel6e^(-)+Cr_2O_7^(2-)to2Cr^(3+)]xx2`
`underline(C_2H_5OHto2CO_2+cancel(12e^(-))`
`underline(C_2H_5OH+2Cr_2O_7^(2-)to4Cr^(3+)+2CO_2)`
For 2 " mol of "`Cr_2O_7^(2-)-=1 " mol of "C_2H_5OH` is required `10xx0.05xx10^(-3) " mol of "Cr_2O_7^(2-)`
`-=(10xx0.05xx10^(-3))/(2) " mol of "C_2H_5OH`
`-=25xx10^(-5)` " mol of "`C_2H_5OH`
`-=25xx10^(-5)xx46g of C_2H_5OH`
Percent of alcohol `-=(25xx10^(-5)xx46)/(10)xx100=0.115%`
Equivalent method:
`Cr_2O_7^(2-)-=C_2H_5OH`
`1mEq-=1mEq`
`10xx0.05xx6m" Eq of "Cr_2O_7^(2-)=3 mEq-=3 m" Eq of "C_2H_5OH`
`3 mEq-=3 mEq`
`3m" Eq of "C_2H_5OH-=3xx10^(-3)eq-=3xx10^(-3)xx(46)/(12)g`
Percent of alcohol `-=(3xx10^(-3)xx46xx100)/(12xx10)-=0.115%`