I have two Chemistry related questions that were in my AP Chem summer packet.
The first has to do with Stoichiometry.
It says, "Calculate the number of moles, molecules, and atoms in 60.0 g of boron triflouride.
I calculated that there are .885 moles, and 5.33 x 10^23 molecules. But I can't figure out how to convert it to atoms. It says the answer is 2.13 x 10^24, but how do I get that?
the second deals with solutions,
It asks, "What mass of NaCl is present in 10.0 g of water when ÄTfus=.206 (Kf= 1.86 °C/m)
The solution says
.206= 1.86 (x/.100)
x= .0110 7(7 =subscript) x 58.44 g/mol = .647 g NaCl
We never covered this in class last year and I don't really understand the process. Could someone explain this to me?
You did the molecules. Aren't there four atoms in borontrifluoride (BF3) per molecule?
On the heat..
The total heat released is dependent on the moles of salt per kg of solution, times a constant.
Totalheat= heat/m *molality
now molality= moles salt/kgwater
=gramssalt/58.44 /0.010
or
totalheat= heat/m*gramssalt/0.010*58.44
gramssalt= .206/1.86 * 58.44*.010
This is a decimal place off your text answer. The difference is here...problems states 10 grams, or .010kg. Your text work uses .100 kg, or 100 grams.
= 1.86C/m*molessalt/.01
For the first question regarding boron trifluoride, you correctly calculated the number of moles and molecules. To convert the number of molecules to the number of atoms, you need to consider the chemical formula of boron trifluoride (BF3). According to the formula, there are four atoms in each molecule: one boron atom and three fluorine atoms.
So, to find the number of atoms, you can multiply the number of molecules (5.33 x 10^23) by the number of atoms per molecule (4).
Calculating this gives us:
(5.33 x 10^23) * 4 = 2.13 x 10^24 atoms
Therefore, the correct number of atoms in 60.0 g of boron trifluoride is 2.13 x 10^24 atoms.
Regarding the second question about the mass of NaCl in water, the process involves using the freezing point depression equation and the freezing point depression constant (Kf).
The equation for freezing point depression is:
ΔTfus = Kf * molality
Where ΔTfus is the change in freezing point, Kf is the freezing point depression constant, and molality is the molal concentration of the solute.
In this case, you are given the freezing point depression (ΔTfus) as 0.206, and the Kf value as 1.86 °C/m.
You set up the equation as follows:
0.206 = 1.86 * (x / 0.100)
Simplifying, you get:
x = 0.206 / 1.86 * 0.100
Next, you multiply the result by the molar mass of NaCl (58.44 g/mol) to get the mass of NaCl.
Therefore, the mass of NaCl in 10.0 g of water is approximately 0.647 g.
It seems that there was a mistake in the calculations in your solution, where you used 0.100 kg instead of 0.010 kg for the mass of water. Taking the correct mass of water into account, the result should be 0.010 instead of 0.100 in the equation and subsequent calculations.
I hope this explanation helps clarify the process.