# chemistry repost!!

The mole fraction of acetic acid (CH3COOH) in an aqueous solution is 0.675. Calculate the molarity of the acetic acid, if the density of the solution is 1.0266 g mL-1.

1. 👍
2. 👎
3. 👁
1. Molarity is moles of solute (acetic acid) per liter of solution. I will call acetic acid X to save typing.

[Moles X/liter solution] = [Moles X/Moles H2O]x [Moles H2O/liter]
=[Moles X/Moles H2O]x[MolesH2O/g H2O][gH2O/g solution][g solution/liter]
=(.325/.675)(1/18)MolesX/g/H2O[gH2O/g solution][1026.6 g/l]
You still need the mass fraction of H2O in the solution. That can be computed from
the mole fractions and molecular masses and turns out to be 0.1644 gH2O/gsolution

1. 👍
2. 👎
2. To give a different take on this, here is another way of approaching the problem. I don't claim it is better or easier. Let X = mols acetic acid; Y = mols H2O. (Two equations and two unknowns).
mol fraction acetic acid = X/(X+Y) = 0.675
solve for X in terms of Y. I have
X = 2.077*Y

(second equation)
mol fraction water = 1-X = 0.325 = Y/(X+Y)
solve for Y in terms of X. I have
0.481*X
Plug 0.481X in for Y and I get 1 mol for X; plug this value into the second equation to obtain Y = 0.481 mol.

Then 1 mol X = 1 mol CH3COOH = 60 g (You should check this as I estimated.)
0.481 mol Y = 0.481 mol H2O = 8.658 g
Total solution = 68.658 g.
Use density to convert this to volume.
Volume = mass/d = 66.879 mL.
M = mols/L = 1 mol/0.066879 = 14.95 M which probably should be rounded to 15 M since it appears we have only 3 significant figures.
Check my thinking. Check my work.

1. 👍
2. 👎
3. wow thanks! its so very clear!

1. 👍
2. 👎

## Similar Questions

1. ### Chemistry

A solution of I2 was standardized with ascorbic acid. Using a 0.1000-g sample of pure ascorbic acid (C6H8C6), 25.32 ml of I2 were required to reach the starch end point. What is the molarity of the iodine solution? Please check to

2. ### Chemistry

An 8.00 mass % aqueous solution of ammonia has a density of 0.9651g/ml. Calculate the molality, molarity, and mole fraction of NH3. This question seems simple but I cannot seem to remember how to do this. Help please?

3. ### Chemistry

Pure acetic acid, known as glacial acetic acid, is a liquid with a density of 1.049 g/mL at 25°C. Calculate the molarity of a solution of acetic acid made by dissolving 35.00 mL of glacial acetic acid at 25°C in enough water to

4. ### Chemistry

You are asked to prepare 500. mL of a 0.150 M acetate buffer at pH 5.00 using only pure acetic acid, 3.00 M NaOH, and water. Calculate the quantities needed for each of the following steps in the buffer preparation. 1. Add acetic

1. ### school

What is the pH of 0.1 M formic acid solution? Ka=1.7„e10-4? What is the pH value of buffer prepared by adding 60 ml of 0.1 M CH3COOH to 40 ml of a solution of 0.1 M CH3COONa? What is the pH value of an acetate buffer (pK=4.76)

2. ### Chemistry

An aqueous antifreeze solution is 40.0% ethylene glycol (C2H6O2) by mass. The density of the solution is 1.05 g/cm3. Calculate the molality, molarity, and mole fraction of the ethylene glycol.

3. ### chemistry

Finding molarity and molality(best answer)? An aqueous solution is prepared by diluting 3.30 mL acetone,(d= 0.789g/mL) with water to a final volume of 75.0 mL. The density of the solution is 0.993 g/mL. What is the molarity,

4. ### chemistry

10 ml sample of vinegar an aqueous solution 0f acetic acid( HC2H3O2) is titrated with 0.5062 M and 16.58 ml is required to reach equivalence point what is the molarity of the acetic acid b. if the density of vinegar is 1.006 g/cm3

1. ### Chemistry

Commercial vinegar was titrated with NaOH solution to determine the content of acetic acid, HC2H3O2. For 20.0 milliliters of the vinegar 26.7 milliliters of 0.600-molar NaOH solution was required. What was the concentration of

2. ### AP Chemistry

An aqueous sulfuric acid solution containing 571.6 g of H2SO4 per liter of solution at 20°C has a density of 1.3294 g/mL. Calculate (a) the molarity, (b) the molality, (c) the percent by mass, and (d) the mole fraction of H2SO4

3. ### Chemistry

Pure acetic acid, known as glacial acetic acid, is a liquid with a density of 1.049 g/mL at 25°C. Calculate the molarity of a solution of acetic acid made by dissolving 10.00 mL of glacial acetic acid at 25°C in enough water to

4. ### chemistry

calculate the molality, molarity, and mole fraction of NH3 in an 12.80 mass % aqueous solution (d=0.9651 g/ml)