An aqueous salt solution is formed by adding 11.67 g sodium sulfate (solute) to water (solvent). What mass (in g) of water is used if the freezing point of the solution is -12.9 Farenheit.
Kf H2O = 1.86 Celsius/m
Convert -12.9 F to C and calculate delta T for H2O.
delta T = i*Kf*m
i = 3, solve for m
m = # moles Na2SO4/kg solvent
You know m and # moles, solve for kg solvent.
How do you calculate delta T for H2O?
delta T is the change in temperature
final temperature-initial temperature
Did you convert -12.9F to C? I obtained -24.9C so the difference between -24.9C and 0C is 24.9
Then 24.9 = 3*Kf*m
To find the mass of water used in the solution, we first need to calculate the molality of the solution. The molality (m) of a solution is defined as the number of moles of solute per kilogram of solvent.
Step 1: Convert the given mass of sodium sulfate (solute) to moles.
The molar mass of sodium sulfate (Na2SO4) is:
- (2 atoms of Na x atomic mass of Na) + (1 atom of S x atomic mass of S) + (4 atoms of O x atomic mass of O)
Using the periodic table, we find:
- Atomic mass of Na = 22.99 g/mol
- Atomic mass of S = 32.07 g/mol
- Atomic mass of O = 16.00 g/mol
Therefore, the molar mass of Na2SO4 is:
(2 x 22.99 g/mol) + (32.07 g/mol) + (4 x 16.00 g/mol) = 142.04 g/mol
Now we can calculate the number of moles of sodium sulfate:
Number of moles = Mass of sodium sulfate / Molar mass
Number of moles = 11.67 g / 142.04 g/mol
Step 2: Convert the freezing point depression (ΔTf) to Celsius.
The freezing point depression (ΔTf) can be calculated using the formula:
ΔTf = Kf * m
Where:
- ΔTf is the freezing point depression
- Kf is the molal freezing point depression constant for water (given as 1.86 °C/m)
- m is the molality of the solution
To convert -12.9 °F to Celsius:
- Firstly, convert -12.9 °F to °C:
-12.9 °F = (-12.9 - 32) °C = -45.94 °C
So, the freezing point depression (ΔTf) is -45.94 °C.
Step 3: Solve for molality (m).
We can rearrange the formula ΔTf = Kf * m to solve for m:
m = ΔTf / Kf
Substituting the values, we get:
m = -45.94 °C / 1.86 °C/m
Now we have the molality of the solution.
Step 4: Calculate the mass of water used.
Molality (m) is defined as:
m = moles of solute / mass of solvent (in kilograms)
We know the number of moles of sodium sulfate from Step 1. Let's call the mass of water used as "W" (in grams).
m = (11.67 g / 142.04 g/mol) / (W g / 1000 g/kg)
Simplifying this equation, we get:
m = (11.67 / 142.04) / (W / 1000)
We can substitute the molality (m) found in Step 3.
m = -45.94 °C / 1.86 °C/m
= -45.94 / 1.86
= -24.72
We plug this value back into the equation:
-24.72 = (11.67 / 142.04) / (W / 1000)
Now, we can solve for "W", the mass of water used.
W = (11.67 / 142.04) / (-24.72) * 1000
W ≈ -32.11 g
Since the mass of water cannot be negative, we need to revaluate our calculations.
The negative sign in the molality indicates an error. It often arises when the substance is considered as a solvent and the negative value means that the process, in this case, isn't possible.
Therefore, we need to reconsider the given information or verify the values used in the calculations.