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6. A student titrates 3 samples of KHP. The experimental data collected is shown below. Titration # KHP (g) | Vínal (mL) | Vmital (mL) | Votal NaOH (mL) 20.50 45.f. 1 0.7415 22.05 1.55 2 0.7115 40.80 22.05 18.75 3 0.7273 21.03 1.35 19.68 a) What is the molarity of NaOH using the data from titration 1? Show all steps in

6. A student titrates 3 samples of KHP. The experimental data collected is shown below. Titration # KHP (g) | Vínal (mL) | Vmital (mL) | Votal NaOH (mL) 20.50 45.f. 1 0.7415 22.05 1.55 2 0.7115 40.80 22.05 18.75 3 0.7273 21.03 1.35 19.68 a) What is the molarity of NaOH using the data from titration 1? Show all steps in

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Corrects the ones that are marked wrong(6a, 6e, and 7) completely and SHOW ALL CALCULATIONS: 

6. A student titrates 3 samples of KHP, The experimental data collected is shown below. KHP (g) | Vfinal (mL) | Vmtal (mL) | Votal NAOH (mL) 0.7415 Titration # 45.f.. 1 22.05 1.55 20.50 0.7115 40.80 22.05 18.75 3 0.7273 21.03 1.35 19.68 a) What is the molarity of NaOH using the data from titration 1? Show all steps in calculation. Moles of KHP =.74154 204.27.9/mol 3.43.10-0ml 70.50ML-10-8nL/L %3D Cuncentration = b) What is the molarity of NaOH using the data from Titration 2? Show all steps in the calculation. Moles of kHP = 0,71154 - 204.14 5/hol =4.48x102mol 3. 68.10 -2 1*uol 75 mL•10-L/L = 0.184 M Concentration = c) What is the molarity of NaOH using the data from Titration 3? Show all steps in the calculation. Moles of KHP 0.7213 4 204.24 ニ 2.510.10-5, Concentintion= Th:08 L10 vnL/L = 0.181M %3D Th.08 m L•10"mL/L d) What is the average molarity? 184+(
Transcribed Image Text:6. A student titrates 3 samples of KHP, The experimental data collected is shown below. KHP (g) | Vfinal (mL) | Vmtal (mL) | Votal NAOH (mL) 0.7415 Titration # 45.f.. 1 22.05 1.55 20.50 0.7115 40.80 22.05 18.75 3 0.7273 21.03 1.35 19.68 a) What is the molarity of NaOH using the data from titration 1? Show all steps in calculation. Moles of KHP =.74154 204.27.9/mol 3.43.10-0ml 70.50ML-10-8nL/L %3D Cuncentration = b) What is the molarity of NaOH using the data from Titration 2? Show all steps in the calculation. Moles of kHP = 0,71154 - 204.14 5/hol =4.48x102mol 3. 68.10 -2 1*uol 75 mL•10-L/L = 0.184 M Concentration = c) What is the molarity of NaOH using the data from Titration 3? Show all steps in the calculation. Moles of KHP 0.7213 4 204.24 ニ 2.510.10-5, Concentintion= Th:08 L10 vnL/L = 0.181M %3D Th.08 m L•10"mL/L d) What is the average molarity? 184+(
What is the relative standard deviation of the molarities? Show all steps in the caleulation. Standurd Deviation = S = 2(Xi-X)- %3D S= 0.177-0.181) M+ (0-184 - 0.181)-M +(0.181-0.181 M = 4.528 ×10- M X 1000 Relative Standard eviation = 4.528×10°M 100 D.181 M 0.004528M. 0.181 M X 100 =2-500%0 Y Is the data within the precision standard?, No, the data is nit within, the precision standard since the fint valve, 0.177 is not, close' to the seçond valve, 0.184. Even though the relative standard devintion is ouly 2.50%,the 4truth is that it can be improved. 8. Should the student perform more titrations? order to reduce even 'more +hote tritrations And thercfore, the relative rtandard deviation of the wmorality of NaCH. Yes the student sliould perfurm, more or der to reduce even 'more the standard deviation 48
Transcribed Image Text:What is the relative standard deviation of the molarities? Show all steps in the caleulation. Standurd Deviation = S = 2(Xi-X)- %3D S= 0.177-0.181) M+ (0-184 - 0.181)-M +(0.181-0.181 M = 4.528 ×10- M X 1000 Relative Standard eviation = 4.528×10°M 100 D.181 M 0.004528M. 0.181 M X 100 =2-500%0 Y Is the data within the precision standard?, No, the data is nit within, the precision standard since the fint valve, 0.177 is not, close' to the seçond valve, 0.184. Even though the relative standard devintion is ouly 2.50%,the 4truth is that it can be improved. 8. Should the student perform more titrations? order to reduce even 'more +hote tritrations And thercfore, the relative rtandard deviation of the wmorality of NaCH. Yes the student sliould perfurm, more or der to reduce even 'more the standard deviation 48
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