28.00 g 0.05961 M Cr2O7^2- 35.46 mL C2H5OH mass %

28.00 g 0.05961 M Cr2O7^2- 35.46 mL C2H5OH mass %

 

 

A person’s blood alcohol (C2H5OH) level can be determined

by titrating a sample of blood plasma with potassium dichromate solution.

 

The balanced equation is:

16H^+(aq) + 2Cr2O7^2-(aq) + C2H5OH(aq) → 4Cr^3+(aq) + 2CO2(g) + 11H2O(l)

 

If 35.46 mL of 0.05961 M Cr2O7^2- is required to titrate 28.00 g of plasma,

what is the mass percent of alcohol in the blood?

 

 

혈장 28.00 g을 적정하는데 0.05961 M Cr2O7^2- 35.46 mL가 필요하다면

혈액 속의 알코올의 질량 백분율(%)은 얼마인가?

 

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(0.05961 mol/L) (35.46/1000 L) = 0.002114 mol Cr2O7^2-

 

 

 

Cr2O7^2- : C2H5OH = 2 : 1 계수비(= 몰수비) 이므로,

반응한 C2H5OH의 몰수 = 0.002114 / 2 = 0.001057 mol

 

 

 

C2H5OH의 몰질량 = 46.07 g/mol

0.001057 mol × (46.07 g/mol) = 0.04870 g C2H5OH

 

 

 

(0.04870 / 28.00) × 100 = 0.1739% C2H5OH