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

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?

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

A 10.69% (weight/weight) Fe(NO3)3
(241.86 g/mol) solution has a density of 1.059g solution/mL
solution. Calculate:
a.) The molar concentration of
Fe(NO3)3 in this solution.
b.) The mass in grams of Fe(NO3)3
contained in one liter of this solution.
c.) The molar concentration of NO3- in
the solution.

A 20.0 wt% solution of CaCl2 (110.98 g/mol) has a density of
1.185 g/mL. What is the mass (in milligrams) of a 19.9-mL solution
of 20.0 wt% CaCl2?What is the mass (in grams) of CaCl2 in 419.4 mL
of a 20.0 wt% solution of CaCl2? What is the formal concentration
of CaCl2 (in molarity) of the 419.4-mL CaCl2 solution described in
the previous question?

A water solution contains 40.0% ethylene glycol (C2H6O2) by mass
and has a density of 1.05 g/mL.
1.) how many grams of ethylene glycol are there in 1.250 L of
solution?
2.) What is the mass of 1.250 L of this solution (grams)?
3.) What is the concentration of this solution in ppm?
4.) What is the molality of the C2H6O2 solution?

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