Chapter 14 1) The solubility of ammonium chloride at 70. C is 60. g of solute pe

Question

Chapter 14

1) The solubility of ammonium chloride at 70. C is 60. g of solute per 100. g of water. Which solution would be saturated at 70. C? a) 90. g of solute in 180. g of water b) 90 .g of solute in 200. g of water c) 30. g of solute in 50. g of water d) 35 g of solute in 50. g of water

2) The solubility of ammonium chloride at 70. C is 60. g of solute per 100. g of water. Which solution would be unsaturated at 70. C? a) 110. g of solute per 200. g of water b) 120. g of solute per 200. g of water c) 30. g of solute per 50. g of water d) 35 g of solute per 50. g of water

3) At which temperature would CO2 gas be most soluble? a) 10. C b) 20. C c) 30. C d) 40. C

4) At which temperature would KCl solid be most soluble? a) 10. C b) 20. C c) 30. C d) 40. C

5) At which temperature would KNO3 solid be most soluble? a) 283 K b) 293 K c) 303 K d) 313 K

6) In a ___ solution, the dissolved solute is in equilibrium with the undissolved solute. a) saturated b) unsaturated c) supersaturated

7) One liter of 2.0 M KCl solution and two liters of 1.0 M KCl solution have the same a) density. b) concentration. c) volume. d) number of moles of solute.

8) What is the percent by mass of the solution formed when 5.0 g of solute is dissolved in 40. g of water? a) 8.0 % b) 11 % c) 13 % d) 14 %

9) What is the percent by mass of a solution formed by dissolving 50.0 g of glucose in 1000. g of water? a) 4.76 % b) 5.00 % c) 5.26 % d) 50.0 %

10) How many moles of solute are dissolved in 400. mL of a 0.200 M solution? a) 0.0800 mol b) 0.800 mol c) 8.00 mol d) 80.0 mo

Explanation / Answer

1) given solubility = 60 / 100 = 0.6

now

a) solubility = 90 / 180 = 0.5

b) solubility = 90 / 200 = 0.45

c) solubility = 30 / 50 = 0.6

d) solubility = 35 / 50 = 0.70

so

the saturated solution is c)

2)

given solubility = 60 / 100 = 0.6

now

a) solubility = 110 / 200 = 0.55

b) solubility = 120 / 200 = 0.60

c) solubility = 30 / 50 = 0.6

d) solubility = 35 / 50 = 0.70

so

the unsaturated solution is a)

3) a) 10 C

4) d) 40 C

5) d) 313 K

6) a) saturated

7) we know that

moles = molarity x volume

so

moles = 2 x 1 = 2 = 1 x 2

so

both the solution have same moles of solute

so

the answer is d) number of moles of solute

8)
we know that

mass percent = mass of solute x 100 / mass of solution

so

mass percent = 5 x 100 / 45

mass percent = 11.11

so

the answer is b) 11 %

9) we know that

mass percent = mass of solute x 100 / mass of solution

so

mass percent = 50 x 100 / 1050

mass percent = 4.76

so

the answer is a) 4.76 %

10) we know that

moles = molairty x volume (L)

so

moles = 0.2 x 0.4

moles = 0.8

so

the answer is option b) 0.8 mol

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