Hi I am doing a lab report and I am having trouble doing these calculations. I k

Question

Hi I am doing a lab report and I am having trouble doing these calculations. I know that the paper tells me how to do it but I am a little confused how to solve it.
I need help with C.7, C.9, C.10, and C.11. i appreciate the help!

Figure 133 During a tiiraion alinkan in Calculations section step Procedure Check when Calodaie vokmt otNeoi ustu neutralize the vineia ihe nitial vonume samples of vine average volume of Nuoit used Total the volumes of Naolt used, and divide by the nanber of satiple you used. volume olume (21 Volume (3) Convert the volume (average NaoH HCSU used to a vohme in liters (LX L (C 5 V1000 Average volunte (int.) NaoH ml, 42 14. .042M Calculate the moles of NaOH using the volume (LI and molarity of the Nao Males NaOH C6 C2 L- 04 NaoH Moles of Moles NaOH used L NaoH used Record the moles of acid present in the vinegar, which Me equal to the moles of NaOH used, Moles H H o, -AT Moks:0, I Moles HC H 0, es Naoll used C.9 1 Calculate the molarity (MJ of the acetic acid (HC H1O2) in the vinegar sample. Malarity (C7)0.00s L 00050 L vinegar

Explanation / Answer

Lab report

C7. moles of NaOH = molarity of NaOH (M) x volume of NaOH used (L)

Taking values and feeding to the given equation,

moles of NaOH = 0.1 M x 0.04294 L

                          = 0.004294 mol

C9. moles of acetic acid = moles of NaOH

So,

moles of acetic acid = 0.004924 mols

molarity of acetic acid = moles of acetic acid/volume of intial vinegar solution

                                    = 0.004294 mol/0.005 L

                                    = 0.8588 M

C10. moles = grams/molar mass

So,

grams of acetic acid = moles of acetic acid x molar mass of acetic acid

                                 = 0.004294 mol x 60 g/mol

                                 = 0.26 g

C11. %(w/v) = mass of solute/volume of solution

So,

percent acetic acid in vinegar = 0.26 g x 100/5 ml

                                                = 5.2%

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