You complete your titrations and find that the concentration of acetic acid in the diluted vinegar solution is 0.023 M. If your dilution step involved taking 25.00 mL of the orginal vinegar solution and diluting this to 250.00 mL with water to create your diluted solution for titration, calculate the concentration of your original vinegar solution.Question 4Answera.0.57b.2.30c.0.00023d.0.23e.0.0023

In this experiment, you will take a 25.00 mL aliquot of vinegar and dilute it to 250.0 mL.  You will then take a 25.00 mL aliquot from this diluted vinegar solution and titrate it against the standardised sodium hydroxide. The titration between acetic acid and sodium hydroxide is a 1:1 stoichiometry.    If your standardised sodium hydroxide solution was determined to be 0.051 M, and it required an average titre (titration volume) of 25.3 mL, what is the concentration (in M) of the undiluted vinegar sample (the initial vinegar sample)?  Provide your answer to 2 significant figures.

Measurements for the titration of acetic acid with NaOHEnter your measurements from lab for trails 1 - 3 with the correct number of decimal places  Trial 1Trial 2Trial 3Mass of empty flask (grams)71.39171.42572.095Mass of flask + vinegar (grams)81.00381.37982.014Volume of vinegar (L)0.010000.010000.01000Concentration of NaOH (mol/L)0.20000.20.2Initial volume of NaOH (mL)0.600.290.20Final volume of NaOH (mL)41.6541.9441.53AttendanceThis QR code needs to be scanned by your TA to confirm that you have completed data collection and cleaned up your workspace.Done online(51pts) CalculationsTitration of Acetic Acid with NaOHCalculations for the titration of acetic acid with NaOHTable view List viewComplete the calculations below for trials 1 - 3 with the correct number of decimal places  Trial 1Trial 2Trial 3Mass of empty flask (grams)71.39171.42572.095Mass of flask + vinegar (grams)81.00381.37982.014Mass of vinegar (grams)Volume of vinegar (L)0.010000.010000.01000Concentration of NaOH (mol/L)0.20000.20000.2000Initial volume of NaOH (mL)0.600.290.20Final volume of NaOH (mL)41.6541.9441.53Volume of NaOH added (mL)Volume of NaOH added (L)Moles of NaOH used (mol)Moles of acetic acid (mol)Mass of acetic acid (grams)Mass percent of acetic acid(9pts)Average mass percent(9pts)Standard Deviation

On domestic products such as vinegar, the concentration of active ingredients is rarely reported in Molar units.  It is more common to report the concentration as a composition such as mass per 100 mL or volume per 100 mL.  Vinegar is usually reported as a volume composition (volume of acetic acid per 100 mL vinegar - written as % v/v).  To convert a molar concentration to % v/v:Calculate how many moles of acetic acid is in 100 mL of solution (Hint: n = C x v)Calculate the mass of acetic acid from the amount of moles determined in step 1 (Hint: m= MW x n)Calculate the volume that the mass determined in step 2 would occupy.  To do this, you will have to divide the mass (determined in step 2) by the density of acetic acid (1.049 g.mL⁻¹).  The density states how much 1 mL of a liquid sample weighs - for acetic acid, 1 mL weighs 1.049 g.The answer in step 3 states what volume of acetic acid is present in a 100 mL sample of vinegar which equates to the composition we are after.  So we can just add "% v/v" to the end of our value to denote this is the volume of acetic acid per 100 mL of vinegar. If, from titration, you determined that concentration of acetic acid in vinegar is 0.928 M, what is the composition of acetic acid in vinegar (% v/v).  Answer to 3 significant figures (please enter numeric value only)

Which solution will be in the burette for the first set of titrations (standardisation of your diluted NaOH)?Question 2Answera.diluted NaOH solutionb.diluted vinegar solutionc.distilled waterd.diluted HCl solutione.KHP solution

A student found a 500 cm3 sample of vinegar. Knowing that vinegar is a solution of ethanoic acid (CH3COOH) and water, this student decided to take a 25.0 cm3 sample, dilute it with 25.0 cm3 of water and then react it with sodium hydroxide. The dilute sample required 37.0 cm3 of 0.850 mol dm–3 sodium hydroxide solution to react entirely with it via the following reaction:                CH3COOH(aq) + NaOH(aq) → NaCH3COO(aq) + H2O(l)Calculate the concentration of ethanoic acid in the original 500 cm3 sample of vinegar.