I will appreciate any help. Thanks Results Molarity of the NaOH solution (on the

Question

I will appreciate any help. Thanks

Results Molarity of the NaOH solution (on the bottle): -0.25M Titration Number 43.1 ml40.1l 37.35 ml 3 4 (if needed) Final buret reading (mL) Initial buret reading (mL) 9.30 ml6.10 ml 2.89 ml Volume of NaOH used (mL) 33.8 ml 34.0 m34.46 ml Moles of NaOH used 0.00845 0.00850.008615 --| 0.00845 | 0.0085 | 0.008615 mol mol mol Moles of HC H.O, used mol mol mol Molarity of HC;HO Mc) 0.0845 M 0.085M 0.08615M Average concentration (M): 0.08522 M Questions Calculate and record the number of moles of NaOH used for the two titrations within 0.3 mL of each other. 1. 2. Calculate the moles of the vinegar (HC2H:O2) from each of the two titrations. 3. Calculate the Molarity of the vinegar (HC2H,02) from each of the two titrations and record in the table. (See Introduction) 4. Calculate the average molarity. 5. Determine the grams of HC2H,0, from the average moles. 6. The original bottle of vinegar claims that the vinegar contains 5% acetic acid by mass. Calculate the mass % using the data from your experiment. Compare your result to what is written on the bottle.

Explanation / Answer

Please correct your initial and final reading in the table. Initial reading should be higher than your final reading which is not so in your case. Please interchange your values.

Another point to note is, check your burette, it may not be as precise as second decimal place. But the values reported is shown to be precise till second decimal place.

1) We know,

Moles of NaOH used = Molarity of NaOH x Volume of NaOH

Expt. 1: Moles of NaOH used = 0.25 x 33.8 = 8.45 milli moles

Expt. 2: Moles of NaOH used = 0.25 x 34.0 = 8.5 milli moles

Expt. 3: Moles of NaOH used = 0.25 x 34.46 = 8.615 milli moles

2) Moles of vinegar used = Moles of NaOH used up during titration

Expt. 1: Moles of Vinegar used = 0.00845 moles

Expt. 2: Moles of Vinegar used = 0.0085 moles

Expt. 3: Moles of Vinegar used = 0.008615 moles

3) Volume of vinegar solution used is not given in your question. But seeing your result it seems that the volume of the vinegar solution taken is 100 ml.

We know, Molarity = No. of moles/Volume of solution

Expt. 1: Molarity of Vinegar solution = 0.00845/0.100 M = 0.0845 M

Expt. 2: Molarity of Vinegar solution = 0.0085/0.100 M = 0.085 M

Expt. 3: Molarity of Vinegar solution = 0.008615/0.100 M = 0.08615 M

4) Therfore average molarity= (M1 +M2 +M3)/3 = (0.0845 + 0.085 +0.08615)/3 = 0.0852 M

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