Question

Please help break down

1. Molaroty unknown, mmol/ml

2. Average molarity of unknown

3. Relative average deviation

20.000ル 20.00ル 26001c volume of NaOH used, mL Molarity of NaOH, mmol/ml Average molarity of unknown Relative average deviation B、Titration of Unknown Acid Number of unknown acid: O.990 Molarity of NaOH: (average from Part A) Final -nha Trial 1 Trial 4 A Initial buret reading, ml ゆFinal buret reading, mL 0.00 mし000ML LOON 22-V0 ML aa·3DM.23.GHU C Volume of NaOH used, mL. 20.00 nu 20.0 D Volume of unknown used, mL 22.wn. 12.30 m 21.4 Mし Molarity of unknown, mmol/mL Average molarity of unknown Relative average deviation Sample Calculations

Answer 1

V_acid*M_acid = V_NaOH *M_NaOH (1)

Trial 1

V_NaOH = 20.00 ml. M_NaOH = 0.990 M

V_acid = 22.60 ml, M_acid = X M

using Eq.1

20.00*0.990 = 22.60*X

or, X = (20.0*0.990) /22.60 = 0.8761 M = 876.1 mmol/ L

Trial 2

V_NaOH = 20.00 ml , M_NaOH = 0.990

V_acid  = 22.30 ml , M_{acid =} X M

using Eq.1

20.00*0.990 = 22.30*X

or, X = (20.00*0.990)/22.30 = 0.8878 M = 887.8 mmol/ L

Trial 3

V_NaOH = 20.00 ml, M_NaOH = 0.990 M

V_{acid =} 22.60 ml, M_acid = X M

Using Eq.1

20.00*0.990 = 22.60*X

or, X = (20.00*0.990)/22.60 = 0.8761 M = 876.1 mmol/ L

hence, Average molarity ( $\mu$)

= ( 0.8761 + 0.8878 + 0.8761)/3 = 0.88 M = 880 mmol/ L

Now, average deviation = $\sum ( |\mu - x)|/n$

$\mu$ is average, x is data value, n is number of trials

= (|0.88 - 0.8761|)+(|0.88 - 0.8878|) +(|0.88 - 0.8761|) /3

= (0.0039+0.0078+0.0039)/3

= 0.0052

Relative average deviation = average deviation *100 = 0.0052*100 = 0.52