0.10% 50.0 g sample 0.0723 M K2Cr2O7 38.94 mL

0.10% 50.0 g sample 0.0723 M K2Cr2O7 38.94 mL

 

 

Laws passed in some states define a drunk driver

as one who drives with a blood alcohol level

of 0.10% by mass or higher.

The level of alcohol can be determined by titrating blood plasma

with potassium dichromate according to the following equation

16H^+(aq) + 2Cr2O7^2-(aq) + C2H5OH(aq)

→ 4Cr^3+(aq) + 2CO2(g) + 11H2O(l)

 

Assuming that the only substance that reacts with

dichromate in blood plasma is alcohol,

is a person legally drunk

if 38.94 mL of 0.0723 M potassium dichromate is required

to titrate a 50.0 g sample of blood plasma?

 

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반응한 K2Cr2O7의 몰수

= (0.0723 mol/L) (38.94/1000 L) = 0.002815 mol K2Cr2O7

( 참고 https://ywpop.tistory.com/7787 )

 

 

 

에탄올 산화 반응식

2Cr2O7^2-(aq) + C2H5OH(l) + 16H^+(aq)

→ 4Cr^3+(aq) + 2CO2(g) + 11H2O(l)

( 참고 https://ywpop.tistory.com/7326 )

 

 

 

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

0.002815 mol K2Cr2O7과 반응한 에탄올의 몰수를 계산하면,

Cr2O7^2- : C2H5OH = 2 : 1 = 0.002815 mol : ? mol

 

? = 0.002815 / 2 = 0.0014075 mol C2H5OH

 

 

 

C2H5OH의 몰질량 = 46.07 g/mol 이므로,

0.0014075 mol × (46.07 g/mol) = 0.06484 g C2H5OH

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

에탄올의 질량 백분율을 계산하면,

(0.06484 g / 50.0 g) × 100 = 0.130%

---> 만취 운전자이다.

 

is a person legally drunk? ---> Yes.

 

 

 

 

[키워드] 만취 운전자 기준문서, 음주 운전자 기준문서, 음주 운전 기준문서