My professor gave us two homework questions that are a bit confusing, can you help me to get started on them.
1) You have 6g of water (Kf=1.86 C kg/mol). You add and dissolve 2.1 g of a different unknown substance. The freezing point of the solution is -0.3 deg C. What is the molecular mass of the solute?
2) A nitric acid solution has a density of 1.42g/ml and is 16 molar. What is its molality? What is the mole fraction of nitric acid? What is the weight percentage?
If you could help me out I would greatly appreciate it.
#1. This type problem illustrates the colligative properties of H2O. The freezing point of pure water is zero degrees C but a solute added makes the solution freeze lower than zero C.
delta T = Kf*molality
solve for molality.
Then molality = moles/kg solvent.
solve for moles.
Then moles = g/molar mass
solve for molar mass.
#2. The HNO3 solution is 16 M. That means 16 mols HNO3/L of solution. The solution has a density of 1.42 g/mL; therefore, 1000 mL has a mass of
1.42g/mL x 1000 mL = 1420 g.
Part of that is water and part is HNO3. You know 16 moles of it is HNO3, the molar mass of HNO3 is approximately 63 (you should do it more accurately) and 16 moles would be 1008g; the amount of water must be 1420-1008 = ??
Then molality = moles/kg solvent.
Post your work if you get stuck.
I did both problems and these are my results:
1) Molality concentration= -.161 kg/mol
8.1g--> .0081 kg
mols solute: (-.161kg/mol)(.0081kg)= -.0013 mol
then, -.0013 mol H2O(18.016 g H2O)= -.0234 g H2O solute
2) HNO3= 16 mols/l
density= 1420 g
16 mols HNO3(63.015)= 1008.2 g HNO3
1420-1008= 412g
Molality: 412g/1000g=.412 kg
then, 16 mols/.412 kg= 38.8= molality
mol fraction: (1.42g)(1000ml)=1420 g/l
mols solvent: (1420g)/(63.015g/mol)= 22.5 mols solvent
total # mols solution:
=16 mol+22.5 mol=38.5 mols solution
mol fraction: 16 mol/38.5= .4155
weight percentage: mass solute/mass solution *100%
38.5 mols*63.015g= 2426 g HNO3
so, (1008g/2426g)*100%= 41.5% HNO3
these are my solutions, can you check to see I didn't make any mistakes, thank you
Somewhere along the way you got mixed up on #1. The 0.161 m is correct but you go downhill from there.
1) Molality concentration= -.161 kg/mol
8.1g--> .0081 kg
mols solute: (-.161kg/mol)(.0081kg)= -.0013 mol
then, -.0013 mol H2O(18.016 g H2O)= -.0234 g H2O soluteI don't know why you are working on water---H2O is the solvent, not solute.
0.161 m = moles/kg solvent.
Kg solvent is 0.006 (6 g from the problem) so moles = 0.161/0.006 = 9.67 x 10^-4
and moles = g/molar mass;
molar mass = g/moles = 2.1/9.67 x 10^-4 = ??
2) HNO3= 16 mols/l
density= 1420 g
16 mols HNO3(63.015)= 1008.2 g HNO3
1420-1008= 412g
Molality: 412g/1000g=.412 kg
then, 16 mols/.412 kg= 38.8= molality
OK to here.
mol fraction: (1.42g)(1000ml)=1420 g/l
mols solvent: (1420g)/(63.015g/mol)= 22.5 mols solvent
The 1420 grams is the MASS of the SOLUTION (not the solvent). You worked the molality and calculated 0.412 kg solute above. You should have used 412 g solute here. I think if you correct that the rest will fall into line although you may need to recalculate each step
total # mols solution:
=16 mol+22.5 mol=38.5 mols solution
mol fraction: 16 mol/38.5= .4155
weight percentage: mass solute/mass solution *100%
38.5 mols*63.015g= 2426 g HNO3
so, (1008g/2426g)*100%= 41.5% HNO3
Of course, I'd be happy to help you get started on these questions!
1) To find the molecular mass of the solute, we can use the formula:
ΔT = Kf * m
where ΔT is the freezing point depression, Kf is the molal freezing point depression constant, and m is the molality of the solution.
First, we need to calculate the molality (m) of the solution. We know that the mass of water is 6g and the mass of the solute is 2.1g. The molar mass of water is approximately 18 g/mol. Therefore, the moles of water is given by:
moles of water = mass of water / molar mass of water
= 6g / 18 g/mol
= 0.333 mol
Similarly, for the solute:
moles of solute = mass of solute / molar mass of solute
Now, we can calculate the molality (m):
molality (m) = moles of solute / mass of water (in kg)
Next, we need to calculate the freezing point depression (ΔT). We are given that the freezing point of the solution is -0.3°C, and the freezing point depression constant (Kf) is 1.86°C kg/mol.
Now, we can rearrange the freezing point depression equation to solve for the molar mass of the solute:
molar mass of solute = (ΔT / Kf) * m
Plug in the values for ΔT, Kf, and m into the equation, and solve for the molar mass of the solute.
2) Let's address the questions one by one:
a) To find the molality of the nitric acid solution, we need to know the number of moles of solute (nitric acid) per kilogram of solvent (water). However, we are only given the density of the solution and its molarity.
First, we need to find the mass of the solution. The density of the solution is 1.42 g/mL. If we assume we have 1 L of the solution, the mass of the solution is:
mass of solution = density * volume
= 1.42 g/mL * 1000 mL
= 1420 g
Next, we need to convert the mass of the solution to kilograms:
mass of solution (kg) = mass of solution (g) / 1000
Now, we can find the number of moles of solute (nitric acid):
moles of nitric acid = molarity * volume of solution (in L)
Since the molarity of the solution is 16 M, and we assumed we have 1 L of the solution, the number of moles of nitric acid is 16 moles.
Finally, we can calculate the molality:
molality = moles of solute / mass of solvent (kg)
Substitute the values and calculate the molality.
b) The mole fraction of nitric acid can be obtained by dividing the number of moles of nitric acid by the total number of moles in the solution.
mole fraction (nitric acid) = moles of nitric acid / total moles in solution
For this question, the total moles in the solution include the moles of nitric acid and any other solute present (if given). In this case, only nitric acid is given, so the total moles in the solution is also 16 moles.
Calculate the mole fraction using the provided equation.
c) To find the weight percentage of nitric acid, we need to know the mass of nitric acid and the mass of the solution.
The mass of nitric acid can be calculated by multiplying the molarity by the molar mass of nitric acid (HNO3), assuming you are given the molar mass.
mass of nitric acid = molarity * molar mass of nitric acid
To find the weight percentage:
weight percentage = (mass of nitric acid / mass of solution) * 100
Substitute the values and calculate the weight percentage.
Remember to double-check all calculations and units to ensure accurate results.