ACID BASE BUFFER PH TITRATION CURVE
WEAK STRONG SOLUTION
STOICHIOMETRY CALCULATION PRACTICE
2026 CHEM 210 EXAM 3
◉ What is the molality of a solution with 12.5 g of ethylene glycol
(C2H6O2) dissolved in 135 g of water?
Answer: 1.49 m
◉ What is the freezing point of a water solution using .85 m
concentration? Freezing point constant is 1.86
Answer: -1.6
◉ What is the Molarity of Nickel II chloride at 0.500 liters with 1
mole of Nickel II chloride solution?
Answer: 2 M
◉ At what point is Copper II sulfate a saturated solution at a volume
of with 0.200 liters?
Answer: 0.150 mol
, ◉ What appears to be forming in the Copper II sulfate solution once
it becomes saturated?
Answer: crystals
◉ Using 1 mole of Gold II chloride, what volume is needed to achieve
a 1 M solution?
Answer: 1.00 L
◉ Could these values be obtained without this simulation? If so,
how? If not, why not?
Answer: Yes, because the formula for Molarity could be used in each
instance.
◉ What happened when salt was added to the ice?
Answer: The freezing point was lowered (freezing point depression).
◉ What differences did the amount of salt used make?
Answer: It affected how long it took the ice cream to freeze. and It
affected the temperature of the water.
◉ What is another use for salt?
Answer: Use on snowy roads.
WEAK STRONG SOLUTION
STOICHIOMETRY CALCULATION PRACTICE
2026 CHEM 210 EXAM 3
◉ What is the molality of a solution with 12.5 g of ethylene glycol
(C2H6O2) dissolved in 135 g of water?
Answer: 1.49 m
◉ What is the freezing point of a water solution using .85 m
concentration? Freezing point constant is 1.86
Answer: -1.6
◉ What is the Molarity of Nickel II chloride at 0.500 liters with 1
mole of Nickel II chloride solution?
Answer: 2 M
◉ At what point is Copper II sulfate a saturated solution at a volume
of with 0.200 liters?
Answer: 0.150 mol
, ◉ What appears to be forming in the Copper II sulfate solution once
it becomes saturated?
Answer: crystals
◉ Using 1 mole of Gold II chloride, what volume is needed to achieve
a 1 M solution?
Answer: 1.00 L
◉ Could these values be obtained without this simulation? If so,
how? If not, why not?
Answer: Yes, because the formula for Molarity could be used in each
instance.
◉ What happened when salt was added to the ice?
Answer: The freezing point was lowered (freezing point depression).
◉ What differences did the amount of salt used make?
Answer: It affected how long it took the ice cream to freeze. and It
affected the temperature of the water.
◉ What is another use for salt?
Answer: Use on snowy roads.
