can someone help me with these questions please. VBNM palanted eylinler beforeha

Question

can someone help me with these questions please. VBNM palanted eylinler beforehand and found the mass to be 30.214g. The combined masof vinegar with the graduated cylinder came to 56472 What is the mass of 25mL of vinegrt What is the density of vinegar in g'ml? You titrate your 25.00mL. sample with 0.9918M NaOH of known concentration. You fill a 50ml buret with NaOH solution and add your vinegar to an Erlenmeyer phenolphthalein. You slowly drip NaOH in to your sample and notice the solution tures pink with each drop added. Upon swirling, the pink color subsides. After some time, you swirl the solution and a pale pink color persists. You read the volume on the buret and determine that 21.Iml of NaOH have been dispensed. You add a few more drops of NaOH and the solution turns a bright pink.(MMt of Acetic Acid 6005 g/mol) flask with a few drops of .What is your analyte? What is your titrant? Which substance serves as the standard solution? What term refers to the point at which the solution retains a pale pink color? How do you know you have reached the end point? Approximately how many What is the percent composition by mass of acetic acid in 25mL of vinegar? Show Consider the following reaction of solid rocket fuel:

Explanation / Answer

1.

Mass of vinegar = (56.472- 30.214) g = 26.258 g

Density of vinegar = mass/ volume = 26.258 g/ 25.0 mL = 1.05 g/ mL

2.

Analyte is: vinegar solution

Titratnt is: NaOH solution

Standard solution: NaOH solution

The point at which the solution retains a pale pink color : equivalence point

Calculation of moles of acetic acid (CH3COOH):

The balanced neutralisation chemical reaction is: CH3COOH + NaOH = CH3COONa + H2O

Mole ratio of CH3COOH : NaOH = 1:1

21.1 mL of 0.9918 M of NaOH is needed to reach the equivalence point. Number of moles of NaOH = (21.1*0.9918)/1000 mole = 0.021 mole

0.021 mole NaOH reacts with 0.021 mole of CH3COOH

Answer: Approximately 0.021 mole of CH3COOH is present in 25 mL. vinegar

Calculation of mass percentage of acetic acid in vinegar:

Moles of acetic acid is 0.021 mole and the molar mass is 60.05 g/ mol

So, the mass of acetic acid = 0.021 mole * 60.05 g/ mole = 1.26 g

Mass of vinegar solution = 26.258 g

Mass percent of acetic acid in vinegar = (1.26/ 26.258)*100% = 4.8%

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