Question

Assuming that the vinegar solution has a density of 1.05 g/mL, determine the percentage (weight/weight) of acetic acid in the vinegar solution. (It has a 0.275 molarity and there are 1.33 Liters of it, or 22.0 grams of acetic acid in it)

Answer #1

concentration of acetic acid in vinegar = 0.275 M

no. of moles of acetic acid in 1L vinegar solution = 0.275 mol

mass of acetic acid (molar mass = 60.05 g/mol) in 1L vinegar solution = (0.275 mol)(60.05 g/mol) = 16.51 g

density of vinegar solution = 1.05 g/mL

mass of 1 L vinegar solution = density x volume = (1.05 g/mL)(1000 mL) =1050 g

Therefore, 1050 g of vinegar solution has 16.51 g acetic acid.

Hence, percentage of aetic acid(w/w) in vinegar solution =
(16.51 g/ 1050g ) x 100 = **1.57 %**

Assuming the density of vinegar is 1.00 g/mL, determine the
average percentage by mass of acetic acid in vinegar. Molarity of
NaOH = .3 Volume of NaOH = 29.2 Mass of Vinegar = .205 mol

) A sample of vinegar is 4.8% acetic acid, HC2H3O2, by weight,
andhas a density of 1.00g/mL. Calculate (a) the grams of acetic
acidin 1 liter of vinegar (b) the molarity of acetic acid
invinegar. 2) calculate the concentration of an unknown acid
solution,HA, if 25.24 mL of .1278 M NaOH solution were needed to
neutralize 28.20 Ml of the HA solution

the density of
a 6.27 M aqueous acetic acid solution is 1.045 g/ml. determine the
molarity of this solution and mole fraction of acetic
acid

Concentrated glacial acetic acid is 99.7% H3CCO2H by mass and
has a density of 1.05 g/mL. What is the molar concentration of
concentrated glacial acetic acid?

The label on a bottle of vinegar indicates that it is 3.5%
acetic acid (CH3COOH). If the density of the solution is
1.90 g/mL, what is the molarity of the solution?

vinegar is an aqueous solution of acetic acid. By law, it must
contain 4 g acetic acid (CH3CO2H 60.05 g/mol) per 100 mL solution
although it is commonly up to 8% acetic acid. If a 15.00 mL aliquot
of vinegar required 5.28mL of 0.10 M NaOH to reach the end point,
is the sample legally vinegar?

A 50.0-mL solution is initially 1.49% MgCl2 by mass and has a
density of 1.05 g/mL.
What is the freezing point of the solution after you add an
additional 1.34 g MgCl2? (Use i = 2.5 for MgCl2.)
Express your answer using two significant figures.

Background info: Vinegar contains acetic acid, CH3COOH. You can
determine the mass of acetic acid in a vinegar sample by titrating
with sodium hydroxide of known concentration. The reaction:
CH3OOH(AQ) +NaOH(aq)=> CH3COONa(aq)+H2O(l). I have determined
the grams of acetic acid to be 60. There are 25 ml sample of
vinegar requiring 41.33 mL of a .953 M solution of NaOH by
titrating sodium hydroxide of known concentration.
** THE QUESTION IS: What is the molar concentration of
the acetic acid...

A 3.455 g sample of vinegar is titrated. It requires 16.83 mL of
0.2145 M NaOH to get to the phenolphthalein end point. Calculate
the weight percent of acetic acid in this sample of vinegar.

Determine the molarity and molality of an aqueous concentrated
hydrochloric acid solution that is 33% by mass acid with a density
= 1.33 g/mL

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