Help me please? Water Mass of Test Tube: 27.4g Volume of Water: 10 g Room Temper
ID: 580281 • Letter: H
Question
Help me please?
Water
Mass of Test Tube:
27.4g
Volume of Water:
10 g
Room Temperature:
21.5 ºC
Freezing Point:
0.0 ºC
FP Sample #1
Mass of FP Sample #1:
2 g
Mass of Test Tube:
27.4g
Volume of Sample:
1.3 mL
Volume of Sample and Water:
11.3 mL
Mass of Sample and Water:
12 g
Room Temperature:
21.5 ºC
Mixture Temperature Decrease:
19.3 ºC
Freezing Point:
-2.1 ºC
FP Sample #2
Mass of FP Sample #2:
2 g
Mass of Test Tube:
27.4g
Volume of Sample:
1.2 mL
Room Temperature:
21.5 ºC
Mixture Temperature Increase:
21.8 ºC
Freezing Point:
-4.6 ºC
Create a table of the following data for BOTH FP samples 1 and 2. Provide a Table title and description. Include units. Any missing information may result in point deductions.
T
Molar Mass (Mm)
Number of moles (n)
Molality (m)
Next, in a paragraph, discuss the results of this experiment presented in this section. Also, identify samples 1 and 2 from the data provided in the table below.
Name of Compound
Molar Mass (g/mol)
calcium sulfate
136.14
2'-Nitrocinnamaldehyde
177.15
glucose
180.16
potassium hydrogen phthalate
204.22
sodium oxalate
134.00
Conclusions (2 points)
In a paragraph, briefly discuss the main findings of this experiment. (Hint: think about what this experiment was used to determine/study and think about the results you discussed in the preceding section) (2 points)
Water
Mass of Test Tube:
27.4g
Volume of Water:
10 g
Room Temperature:
21.5 ºC
Freezing Point:
0.0 ºC
Explanation / Answer
We shall need the freezing point depression constant of water which is Kf = 1.86°C/m. Use the expression for freezing point depression as
Tf = Kf.(molality of solution) where we define the molality of the solution as
molality = (moles of solute)/(kilograms of solvent)
FP Sample #1
FP Sample # 2
Mass of FP Sample# (g)
2.0
2.0
Mass of sample and water (g)
12.0
12.0
Mass of water (g)
12.0 – 2.0 = 10.0
12.0 – 2.0 = 10.0
Mass of water (kg)
(10.0 g)*(1 kg/1000 g) = 0.01 kg
(10.0 g)*(1 kg/1000 g) = 0.01 kg
Depression in freezing point, T = (Freezing point of pure water) – (Freezing point of sample + water) (°C)
(0.0) – (-2.1) = 2.1
(0.0) – (-4.6) = 4.6
Molar mass of the sample, Mm (g/mol)
177.143 (check calculation 1 below)
80.869
Number of moles = (mass of sample)/(Mm) (mole)
(2.0)/(177.143) = 0.01129 0.0113
(2.0)/(80.869) = 0.02473 0.0247
Molality of the solution = (moles of sample)/(kg of solvent) (m)
(0.0113)/(0.01) = 1.13
(0.0247)/(0.01) = 2.47
Identity of the sample
2’-nitrocinnamaldehyde (the calculated molar mass matches the molar mass of 2’-nitrocinnamaldehye within the limits of error)
- (None of the given samples matches the calculated molar mass within the limits of error).
Calculation 1:
Let the molar mass of the sample be M g/mol; the number of moles of the sample = (2.0 g)/(M g/mol) = (2/M) mole.
Molality of the solution = (moles of sample)/(kg of solvent) = (2/M) mole/(0.01 kg) = (200/M) m where m denotes molality of the solution.
As per the expression, we have,
2.1°C = (1.86°C/m)*(200/M) m
====> 2.1°C = 372/M °C
====> M = 372/2.1 = 177.1428 177.143
The molar mass of the FP Sample #1 is 177.143 g/mol (ans).
FP Sample #1
FP Sample # 2
Mass of FP Sample# (g)
2.0
2.0
Mass of sample and water (g)
12.0
12.0
Mass of water (g)
12.0 – 2.0 = 10.0
12.0 – 2.0 = 10.0
Mass of water (kg)
(10.0 g)*(1 kg/1000 g) = 0.01 kg
(10.0 g)*(1 kg/1000 g) = 0.01 kg
Depression in freezing point, T = (Freezing point of pure water) – (Freezing point of sample + water) (°C)
(0.0) – (-2.1) = 2.1
(0.0) – (-4.6) = 4.6
Molar mass of the sample, Mm (g/mol)
177.143 (check calculation 1 below)
80.869
Number of moles = (mass of sample)/(Mm) (mole)
(2.0)/(177.143) = 0.01129 0.0113
(2.0)/(80.869) = 0.02473 0.0247
Molality of the solution = (moles of sample)/(kg of solvent) (m)
(0.0113)/(0.01) = 1.13
(0.0247)/(0.01) = 2.47
Identity of the sample
2’-nitrocinnamaldehyde (the calculated molar mass matches the molar mass of 2’-nitrocinnamaldehye within the limits of error)
- (None of the given samples matches the calculated molar mass within the limits of error).
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