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The true value was 0.120. What was your percent error?

Sagot :

fichoh

The question is incomplete as the measured value isn't given. However, an hypothesized measured value will be used to explain how to calculate the percentage error.

Answer:

2.5%

Explanation:

Percentage error :

[(measured value - True value)/ True value] * 100%

The measured value is the value obtained during an while using a measuring device.

If the measured value is = 0.123

Then, the percent error will be :

[(0.123 - 0.120) / 0.120] * 100%

(0.003 / 0.120) * 100%

0.025 * 100%

= 2.5%

Answer:

For each trial, compute the mol of titrant; (molarity x L) and keep the number of significant figures to 4.

Trial 1: 12.49 mL =

0.0025

mol NaOH

Trial 2: 12.32 mL =  

0.0025

mol NaOH

Trial 3: 11.87 mL =  

0.0024

mol NaOH

Because the moles of titrant should equal the moles of analyte at the equilibrium point, copy these values into the row of the data table marked “mol of analyte.”

Compute the concentration of HCl for all three trials using the formula molarity = mol / L.

Trial 1 (20.61 mL HCl):

0.121

M

Trial 2 (20.06 mL HCl):

0.125

M

Trial 3 (19.67 mL HCl):

0.122

M

Compute the average concentration (molarity) of HCl for all three trials:

0.123

M.

The true value was 0.120. What was your percent error?

2.5

%

Explanation:

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