20.01 Error from technique If you were to overshoot the endpoint by just 1 drop while you were standardizing the NaOH solution, what would be your % error? Assume the actual volume is 20.00 mL and there are exactly 20 drops in 1.00 mL ( find the value of 1 drop) for the sake of this calculation. Y20 = 0.05 %3D % error = |measured value - actual value|/actual value x 100% %3D 20.05ML-20. mx100 =0.25% coml 20. C0ML vill ve as (4) e, our II + TI-30XIS TEXAS INSTRUMENTS SCI/ENG INS 2nd DRG DEL 10* LOG PRB ex ADIC dle STAT EXIT STAT Ablc DATA STATVAR CLEAR SIN- COS-1 TAN- SIN COS TAN EE ( ) 8. RCL 4. 6. STO OFF ON RESET
20.01 Error from technique If you were to overshoot the endpoint by just 1 drop while you were standardizing the NaOH solution, what would be your % error? Assume the actual volume is 20.00 mL and there are exactly 20 drops in 1.00 mL ( find the value of 1 drop) for the sake of this calculation. Y20 = 0.05 %3D % error = |measured value - actual value|/actual value x 100% %3D 20.05ML-20. mx100 =0.25% coml 20. C0ML vill ve as (4) e, our II + TI-30XIS TEXAS INSTRUMENTS SCI/ENG INS 2nd DRG DEL 10* LOG PRB ex ADIC dle STAT EXIT STAT Ablc DATA STATVAR CLEAR SIN- COS-1 TAN- SIN COS TAN EE ( ) 8. RCL 4. 6. STO OFF ON RESET
Chapter5: Errors In Chemical Analyses
Section: Chapter Questions
Problem 5.9QAP
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Is this calculation correct? See image below....
![20.01
Error from technique
If you were to overshoot the endpoint by just 1 drop while you were standardizing the NaOH solution,
what would be your % error? Assume the actual volume is 20.00 mL and there are exactly 20 drops in
1.00 mL ( find the value of 1 drop) for the sake of this calculation.
Y20 = 0.05
%3D
% error = |measured value - actual value|/actual value x 100%
%3D
20.05ML-20. mx100 =0.25%
coml
20. C0ML
vill
ve as
(4)
e, our
II +
TI-30XIS
TEXAS INSTRUMENTS
SCI/ENG
INS
2nd
DRG
DEL
10*
LOG
PRB
ex
ADIC dle
STAT
EXIT STAT
Ablc
DATA
STATVAR
CLEAR
SIN-
COS-1
TAN-
SIN
COS
TAN
EE
(
)
8.
RCL
4.
6.
STO
OFF
ON
RESET](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F88f794be-122d-44d8-bdbd-e59efaff39a0%2F6f48b9df-33cf-4ca3-b9d3-4e50893da85a%2Flo9tvbm.jpeg&w=3840&q=75)
Transcribed Image Text:20.01
Error from technique
If you were to overshoot the endpoint by just 1 drop while you were standardizing the NaOH solution,
what would be your % error? Assume the actual volume is 20.00 mL and there are exactly 20 drops in
1.00 mL ( find the value of 1 drop) for the sake of this calculation.
Y20 = 0.05
%3D
% error = |measured value - actual value|/actual value x 100%
%3D
20.05ML-20. mx100 =0.25%
coml
20. C0ML
vill
ve as
(4)
e, our
II +
TI-30XIS
TEXAS INSTRUMENTS
SCI/ENG
INS
2nd
DRG
DEL
10*
LOG
PRB
ex
ADIC dle
STAT
EXIT STAT
Ablc
DATA
STATVAR
CLEAR
SIN-
COS-1
TAN-
SIN
COS
TAN
EE
(
)
8.
RCL
4.
6.
STO
OFF
ON
RESET
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