1. The solubility of a substance in 100g water is 50g at 25oC. 55 g of the subst
ID: 785787 • Letter: 1
Question
1. The solubility of a substance in 100g water is 50g at 25oC. 55 g of the substance is dissolved in water at 40oC and cooled to 25oC, no crystals appear. The solution is
A. supersaturated
B. saturated
C. unsaturated
2. In a mixture of 25.0g CaCl2, 40.0g LiCl, and 300 g of water , the mole percent of water is:
A. 93
B. 5.3
C. 1.3
3. In the reaction: NiCl2(aq) + 2NaOH(aq) -> Ni(OH)2(s) + 2NaCl(aq) ; How many mililiters of a 0.200M NaOH solution are needed to react with 18.0 ml of a 0.500M NiCl2 solution?
A. 90
B. 0.090
C. 45
D. 180
4. Shells consisting of CaCO3 can be found in deep waters
True
False
5. In the summer, fish in a pond will tend to be found at lower depths in the water
True
False
6. How many grams of KCl are needed to prepare 300.0 mL of a 0.14 M KCl solution?
A. 75
B. 3.1
C. 4.2
Explanation / Answer
1. The solubility of a substance in 100g water is 50g at 25oC.
55 g of the substance is dissolved in water at 40oC and cooled to 25oC, no crystals appear. The solution is
A. supersaturated
B. saturated
C. unsaturated --> since the decrease in Temperature should have formed crystals if it was supersaturated or saturated, therefore, it is unsaturated
2. In a mixture of 25.0g CaCl2, 40.0g LiCl, and 300 g of water , the mole percent of water is:
mol of CaCl2 = mass/MW = 25/(110.98 ) = 0.225265
mol of LiCl = masS/MW = 40/42.394 = 0.9435
mol of water = mass/MW = 300/18 = 16.66
mol frac of water = mol of water / total mol --> 16.66 / (16.66+0.225265+0.9435) = 0.934
A. 93
B. 5.3
C. 1.3
NONE, mole fractions should not have x > 1 values, choose 0.93
3. In the reaction: NiCl2(aq) + 2NaOH(aq) -> Ni(OH)2(s) + 2NaCl(aq) ; How many mililiters of a 0.200M NaOH solution are needed to react with 18.0 ml of a 0.500M NiCl2 solution?
mmol of NiCl2 = MV = 18*0.5 = 9 mmol
mmol of NAOH = 2*9 = 18 mmol
V = mmol/M = 18/0.2 = 90 mL
A. 90
B. 0.090
C. 45
D. 180
4. Shells consisting of CaCO3 can be found in deep waters
True; since CaCO3 solubility decreases, so CA+2 an CO3-2 will be expected in shallow water, but the deeper you go, the more CaCO3 you find
False;
5. In the summer, fish in a pond will tend to be found at lower depths in the water
Since T increases in summer, this decreases the solubility of Gases, so they must search shallow water on the bottom, expect fishes to go deeper
True
False
6. How many grams of KCl are needed to prepare 300.0 mL of a 0.14 M KCl solution?
M = mol/V
mol = MV = (0.14)(0.3) = 0.042 mol of KCl
mass = mol*MW = 0.042 * 74.5513 = 3.13 g
A. 75
B. 3.1
C. 4.2
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