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1. The density of vinegar was determined using a pycnometer. The empty pycnometer weighed 0.495g...

1. The density of vinegar was determined using a pycnometer. The empty pycnometer weighed 0.495g at 23¤celcius. Filled with vinegar, it weighed 0.800g. Filled with water it weighed 0.896g. The density of water at 23¤celsius is 0.9975g/cm3. A 0.825cm cube sample of NaOH was standardized using 0.815cm cube of 0.0969 M potassium hydrogen phthalate (KHP). The standardized solution was then used to titrate 0.250cm3 of the vinegar. The volume of NaOH needed was 2.205cm3.
The stoichiometry of the reaction is C2H4.O2(aq) +Na.OH(aq) = Na(aq) + C2H3O2(aq) + H2O(l).
(i). Calculate the density of the vinegar.
(ii). Find the concentration of acetic acid in the vinegar in moles per litre, grams per litre and percent by weight.
2. What amount of oxalic acid (H2C2.O4.2H2O, MW=126.0g/mol) is required to prepare 10cm3 of 0.130 M oxalic acid solution?

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Answers (1)

isaaq
4 days ago
number two (2).
solution.
you can't use C1V1 = C2V2 because you were not given the conc of the starting material unlike in the first question where you were given 50% wt/wt.
let's take a long route so you can understand it. you want to prepare 10mL (0.01 L) of 0.13 M oxalic acid. what is the mol of what you want prepare?
recall that Molar conc = no of mol/volume(L).
therefore no of mol of what you want to prepare = molar conc x volume(L)
= 0.130 M x 0.01 L
= 0.0013 mol
what amount in gram (mass) of oxalic acid do you need to have in 10 ml (0.01 L) such that the mol will be 0.0013 mol?
this leads us to the relationship of mol and mass.
recall that no of mol = mass(g)/molar mass(g/mol)
therefore the mass will be => no of mol x molar mass(g/mol)
= 0.0013 mol x 126.0 g/mol
= 0.1638 g
thus you need to weigh out 0.1638 g of the oxalic into a 10 ml flask/container and make up the volume to the 10mL mark.
in a nutshell amount (g) you will need = molar conc of what you to prepare x the volume in L of what you want prepare x molar mass of the compound...
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