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.
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.
i understand everything you put there EXCEPT solving for x.
i get how you get X = 2.077*Y, but i don't understand how if you plug in 0.481X in for Y to solve for x.
x = 2.077(0.481x)
x = 0.999x
then what???
i know this is probably a simple stupid math question, but i need to do another problem like this and i'm stuck at this part...
To solve for X in terms of Y, we can rearrange the equation "mol fraction acetic acid = X/(X+Y) = 0.675" and solve for X. Here's how:
1. Start with the equation: X/(X+Y) = 0.675
2. Multiply both sides by (X+Y) to eliminate the denominator: X = 0.675(X+Y)
3. Distribute the 0.675 to both terms inside the parentheses: X = 0.675X + 0.675Y
4. Subtract 0.675X from both sides of the equation: X - 0.675X = 0.675Y
5. Simplify the left side: 0.325X = 0.675Y
6. Divide both sides by 0.325 to solve for X: X = (0.675Y) / 0.325
So, the expression for X in terms of Y is X = (0.675Y) / 0.325.