Help with Chemistry PLEASE!!!?
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

2 Answers
- Roger the MoleLv 73 weeks ago
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
- Anonymous3 weeks ago
I can only help if you've asked an actual question(which you did not).
can you help me out plz the concentration is 0.101M HCl