Anonymous
Anonymous asked in Science & MathematicsChemistry · 3 weeks ago

Given the volumes can someone tell me how to calculate:

-mmol of HCL

-mmol of NaOH

-Molarity  of NaOH

-Average molarity of NaOH

-Deviation

-Average Deviation Relevance
• In all cases adjust the number of significant digits as you see fit:

mmol of HCl used:

Trial 1: (12.6 mL) x (0.101 M HCl) = 1.2726 mmol

Trial 2: (13.4 mL) x (0.101 M HCl) = 1.3534 mmol

Trial 3: (13 mL) x (0.101 M HCl) = 1.313 mmol

mmol of NaOH:

HCl + NaOH → NaCl + H2O  [balanced as written]

Trial 1: 1.2726 mmolTrial 2:  1.3534 mmolTrial 3:  1.313 mmol

Molarity of NaOH:

Trial 1: 1.2726 mmol / 13.4 mL = 0.09497 mol/LTrial 2:  1.3534 mmol / 14.4 mL = 0.09399 mol/LTrial 3:  1.313 mmol / 13.7 mL = 0.09584 mol/L

Average molarity of NaOH:

all Trials: (0.09497 mol/L + 0.09399 mol/L + 0.09584 mol/L) / 3 =  0.09493

Deviation:

Trial 1:  0.09497 mol/L - 0.09493 mol/L = 0.00004 mol/LTrial 2:  0.09493 mol/L - 0.09399 mol/L = 0.00094 mol/LTrial 3:  0.09584 mol/L - 0.09493 mol/L = 0.00091 mol/L

Average Deviation:

all Trials: ( 0.00004 mol/L +  0.00094 mol/L +  0.00091 mol/L) / 3 = 0.00063 mol/L

• kens3 weeks agoReport

can you help me out plz the concentration is 0.101M HCl

• Anonymous
3 weeks ago

I can only help if you've asked an actual question(which you did not).