A) Remember that molality ( m ) = moles of solute / kg solvent. So, using your d
ID: 544294 • Letter: A
Question
A) Remember that molality (m) = moles of solute / kg solvent. So, using your data from Table 1, calculate the moles of each solute (glycerol, NaCl, and CaCl2) used in this lab by using the experimental molality and the mass (in kg) of water in each trial.. You must show all work to receive credit.
B) Recall that the molar mass of a substance is expressed in g/mol. If we have a mass of a substance, and we know how many moles of it we have in that given mass, we can divide g/mol to determine the molar mass. So, using your experimental data from Table 1 and your answer to the previous question, calculate the molar mass for each solute. You must show all work to receive credit.
C) Calculate the percent error for your experimental values for each solute. Is your percent error value relatively constant with each trial? If so, what might this indicate (include accuracy and precision as part of your answer)? You must show all work to receive credit.
D) What are some possible sources of error in your experiment? Discuss at least 3 in detail.
E) Which solution of salt (NaCl or CaCl2) may be better to salt the roads? Why? (HINT: Don’t just focus on the final freezing point in each case—consider all your data and how it may be tied to efficiency and economic factors.)
Table 1: Freezing Data Point
Solution
Freezing Point (OC)
t (Ti – Tf)
i
Molality (m)
Solute Mass (g)
Solvent Mass (kg)
H2O
-1.3OC
---
---
---
---
---
Glycerol
- 4.4OC
3.1
1
1.67
6.3
0.025
NaCl
-8.6OC
7.3
2
6.79
2.0
0.025
CaCl2
-5.6OC
4.3
2
3.99
2.0
0.025
Table 1: Freezing Data Point
Solution
Freezing Point (OC)
t (Ti – Tf)
i
Molality (m)
Solute Mass (g)
Solvent Mass (kg)
H2O
-1.3OC
---
---
---
---
---
Glycerol
- 4.4OC
3.1
1
1.67
6.3
0.025
NaCl
-8.6OC
7.3
2
6.79
2.0
0.025
CaCl2
-5.6OC
4.3
2
3.99
2.0
0.025
Explanation / Answer
1) From the given data -
The number of mols of glycerol = (molality)(kg of solvent) = (1.67 m)(0.025) = 0.04175 mol
The number of mols of NaCl = (molality) (Kg of solvent) = (6.79 m)(0.025 kg) = 0.169 mol
The number of mols of CaCl2 = (molality) (Kg of solvent) = (3.99 m)(0.025 kg) = 0.9975mol
2) Molar mass of Glycerol = mass / mols = 6.3 g / 0.04175 mol = 151 g/mol
Molar mass of NaCl = mass / mols = 2.0 g /0.169 mol = 11.83 g/mol
Moalr mass of CaCl2 = mass / mols = 2.0 g/0.9975 mol = 2.00 g/mol
3) % error of Glycerol = (experimental molar mass / actual molar mass)*100
= (151 / 92)*100 = 164 %
% error of NaCl = (11.83 / 58.44)*100 = 20.24 %
% error of CaCl2 = (2.00 /111 g) *100 = 1.80 %
D) CaCl2 is better to use in icy roads as it decreases the freezing point of water more than the NaCl.
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