How to get the average deviation for naoh and vinegar solution Choice L Analysis

Question

How to get the average deviation for naoh and vinegar solution
Choice L Analysis of an Unknown Acid sample Standardization of NaOH Titrant Sample 1 Sample 2 Weight of KHP taken 0.634 7 75 3 Initial NaOH buret reading O.OO Final NaOH buret reading 32. 30 2 32.30 36.50 T3 S. 8 Volume NaoH used Moles KHP present 0,003 nal 0.00 33mol o oo Molarity of NaOH solution o.o qCoM o.104 l M o lo 33 M Mean molarity of NaoH solution and average deviation O. fo l M 1. Analysis of vinegar Solution Identification number of vinegar sample used Sample 3 Sample 1 Sample 2 5 m L Sm L Quantity of vinegar taken m O. O O O oo Initial NaoH buret reading 0, O O a Y3.50 4 00 3a. O Final NaOH buret reading a. o o s o 41.00 Volume NaOH used BM Molarity of vinegar o 40 M 0.40M Mean molarity of vinegar and average deviation o M by mass acetic acid present 5.044

Explanation / Answer

Calculation of Average Deviation

Average molarity of Vinegar: (0.80 + 0.90 + 0.83)/3 = 0.84M

average deviation = (|0.80-0.84| + |0.90-0.84| + |0.83-0.84|)/3 = 0.04+0.06+0.01= 0.11/3 =0.036

Average Deviation: 0.036

Average deviation of NaOH for volume are calculated here.

Average volume of Vinegar: (39.00 + 43.50 + 41.00)/3 = 41.16

average deviation = (|39.00-41.46| + |43.50-41.16| + |41.00-41.16|)/3 = 2.46+2.34+0.16= 4.96/3 =1.65

Average Deviation: 1.65

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