PLEASE SHOW PROCEDURE!
You wish to prepare an aqueous solution that has a freezing point of -0.100 degrees Celsius. How many milliliters of 12.0 M HCl would you use to prepare 250.0 mL of such a solution?
[Hint: Note that in a dilute aqueous solution, molality and molarity are essentially numerically equal.]
Your note gives away the solution:Shame on your teacher.
fp=-N*fpdepresssionconst*m
but m=M in dilute..
and for HCl you get two ions per molecule.
-.1=2(-.52)M
solve for M ( I get about .096)
so you want to dilute the HCl from 12 to .096, or dilute it 12/.096=125 times, which means you need one part HCl, 124 parts water
what is one part? 250ml/125parts=2ml
so measure out 248ml water, add 2 ml of the 12M HCl slowly (Always add acid to water).
check my reasoning.
This sounds like solid reasoning, but I apologize, because the answer in the back of my textbook says 0.56 mL of 12.0 M HCl.
I see your error. You used the incorrect freezing point depression constant: .52 instead of 1.86.
To prepare the aqueous solution, you need to determine the amount of solute (12.0 M HCl) needed to achieve the desired freezing point depression.
The formula to calculate the freezing point depression (ΔTf) is given by the equation:
ΔTf = Kf * m * i
Where:
- ΔTf is the change in freezing point
- Kf is the freezing point depression constant for the solvent (for water, Kf = 1.86 °C/m)
- m is the molality of the solute in the solution (which is approximately equal to the molarity for dilute aqueous solutions)
- i is the van't Hoff factor, which represents the number of particles the solute dissociates into when it dissolves (for HCl, i = 2)
In this case, you want the freezing point to be -0.100 °C, and since molality and molarity are essentially equal, you can use molarity directly.
Plug in the values into the equation:
-0.100 °C = 1.86 °C/m * (12.0 M) * 2
Now solve for m:
m = -0.100 °C / (1.86 °C/m * 24.0 M)
m ≈ -0.100 °C / (44.64 °C·M)
m ≈ -0.00224 M·°C
Now, calculate the amount of solute needed using the relationship between molarity, volume, and amount of solute:
Molarity (M) = amount of solute (mol) / volume of solution (L)
Rearrange the equation to solve for the amount of solute:
amount of solute (mol) = Molarity (M) * volume of solution (L)
Since you have the molarity (12.0 M) and volume of the solution (250.0 mL or 0.250 L), you can calculate the amount of solute:
amount of solute = 12.0 M * 0.250 L
amount of solute = 3.0 mol
Therefore, you would need 3.0 moles of HCl to prepare 250.0 mL of the desired solution.