CHEM 103
MODULE 6 2022
Click this link to access the Periodic Table. This may be helpful throughout
the exam.
List and explain if each of the
... [Show More] following solutions conducts an electric
current: sodium chloride (NaCl), hydrochloric acid (HCl) and sugar (C6H12O6).
Sodium chloride (NaCl) is an ionic compound and conducts since it forms ions
in solution. Hydrochloric acid (HCl) is a polar compound and conducts since it
forms ions in solution. Sugar (C6H12O6) is a molecular compound but does not
form ions in solution so it does not conduct.
Explain how and why the presence of a solute affects the boiling point of a
solvent.
The presence of a solute raises the boiling point of a solvent by lowering the
vapor pressure of the solvent. With this lower vapor pressure, more heat (a
higher boiling point) is required to raise the vapor pressure to atmospheric
pressure.
Click this link to access the Periodic Table. This may be helpful throughout
the exam.
Question 4
Not yet graded / 10 pts
Rank and explain how the freezing point of 0.100 m solutions of the following
ionic electrolytes compare, List from lowest freezing point to highest freezing
point.
GaCl3, Al2(SO4)3, NaI, MgCl2
Your Answer:
Al2(SO4)3 -> 2Al
3+
, 3SO4
-2 = = 5 ions -> most ions = lowest freezing
point
GaCl2 -> Ga
3+
, Cl
- = 4 ions
MgCl2 = 3 ions
NaI = 2 ions - > least ions = highest freezing point
GaCl3
3rd lowest FP
→ Ga+3 + 3 Cl
- ∆tf = 1.86 x 0.1 x 4 =
Al2(SO4)3
lowest FP
→ 2 Al
+3 +
3 SO4
-3 ∆tf = 1.86 x 0.1 x 5 =
NaI
= highest FP
→ Na
+ + I
- ∆tf = 1.86 x 0.1 x 2
MgCl2
= 2nd lowest FP
→ Mg+2 + 2 Cl
- ∆tf = 1.86 x 0.1 x 3
FP: Al2(SO4)3 < GaCl3 < MgCl2 < NaI
Click this link to access the Periodic Table. This may be helpful throughout
the exam.
Show the calculation of the mass percent solute in a solution of 20.8 grams
of Ba(NO3)2 in 400 grams of water. Report your answer to 3 significant
figures.
Question 5
Not yet graded / 10 pts
Question 6
Not yet graded / 10 pts
Question 7
Not yet graded / 10 pts
Mass % = (20.8 / 20.8 + 400) x 100 = 4.94%
Click this link to access the Periodic Table. This may be helpful throughout
the exam.
Show the calculation of the molality of a solution made by dissolving 28.5
grams of C8H16O8 in 400 grams of water. Report your answer to 3 significant
figures.
molality = (gsolute / MW) / (gsolvent / 1000)
molality = (28.5 / 240.208) / (400 / 1000) = 0.297 m
Click this link to access the Periodic Table. This may be helpful throughout
the exam.
Show the calculation of the molarity of a solution made by dissolving 35.9
grams of Mg(NO3)2 to make 400 ml of solution. Report your answer to 3
significant figures.
Molarity = (gsolute / MW) / (mlsolvent / 1000)
Molarity = (35.9 / 148.325) / (400 / 1000) = 0.605 M
Click this link to access the Periodic Table. This may be helpful throughout
the exam.
Question 8
Not yet graded / 10 pts
Show the calculation of the mass of Ba(NO3)2 needed to make 250 ml of a
0.200 M solution. Report your answer to 3 significant figures.
Molarity = (moles) / (mlsolvent / 1000)
0.200 = (moles) / (250 / 1000)
Moles = 0.200 x 0.250 = 0.0500
Moles = (gsolute / MW)
0.0500 = (gsolute / 261.55)
gsolute = 0.0500 x 261.55 = 13.1 g
Click this link to access the Periodic Table. This may be helpful throughout
the exam.
Show the calculation of the volume of 0.667 M solution which can [Show Less]