Chem 121 Lab Final Practice Quiz

5. If the label on the bottle lists the citric acid content as 1 g per serving with a serving size of 240 mL determine the percent error in the titration results. Express result with 2 significant figures.

The following volumes of 0.000300 M SCN- are diluted to 15.00 mL. Determine the concentration of SCN- in each sample after dilution. These values will be used during the experiment.  Sample 0.000300 M SCN- (mL) [SCN-] (M) 1  1.50   2 3.50   3 7.00   4 10.00

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Don

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Fill in the blanks. Calculate the Average molarity (M) of Acetic acid in vinegar, using the following data. Calculate the mass percent of acetic acid in vinegar, using your average Acetic acid concentration (M) from the above result. Table 2. Data for titration of 20.00 mL of vinegar with approximately 1.2 M NaOH.

A student performed titration to find the concentration of acetic acid. Titrant was sodium hydroxide. He went beyond the indicator color change by adding too much titrant. How will his experimental error affect the calculated value for the concentration of unknown acetic acid? calculated value actual value. calculated value < actual value calculated value = actual value. more information is necessary this is not a source of error

[References] Use the References to access important values if needed for this question. In the laboratory, a student dilutes 17.8 mL of a 9.01 M hydroiodic acid solution to a total volume of 200.0 mL. What is the concentration of the diluted solution? AM 3 Concentration = Submit Answer M Try Another Version 1 item attempt remaining Cengage Learning Cengage Technical Support Previous

a) Determine the concentration of acetic acid in the 25.00 mL aliquot. b) Determine the concentration of acetic acid in the 10.0mL of vinegar. concentration NaOH = 0.101M Trial 3 Initial NaOH level 35.1 mL Final NaOH level 52.5mL The volume of NaOH used to reach the endpoint 17.4 mL

STARTING AMOUNT X = 0.1 A student dissolves 24.3 g Na2S₂O3 in enough water to make a 0.302 solution. How many mL solution did the student make? ADD FACTOR x( ) 0.302 24.3 509 1 DELETE 6.022 x 10²3 0.509 L solution g Na₂S₂O3 A 158.1 ANSWER 1000 0.001 1.965 0.01 M RESET 5 3.311 100 M Na2S₂O3 kg Na₂S₂O3 mol Na2S₂O3 mL solution 10

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What volume in L of a 0.32M Mg(NO:)2 solution contains 45 g of Mg(NO:): ? X STARTING AMOUNT ADD FACTOR ANSWER RESET *( ) 148.33 86.32 10 45 1 210.34 6.022 x 1023 0.097 0.95 0.32 140 0.30 L mL mol Mg(NOs): M Mg(NOs): g Mg(NOs)2 g Mg(NO:)»/mol