S_00-001

Question 1

Which of the two compounds (H2O or NH3) forms the stronger hydrogen bonds?
Justify

Answer 1

H2O because it has the largest δ+ and δ ‒ because O is more EN than N. The larger the ∆EN is the larger the δ+ and δ ‒ are and the stronger the attraction is

Question 2

What is entropy, what causes entropy

Answer 2

The dispersion of particles (in a volume) is called entropy and results in the
dispersion of the energy they contain.
(Kinetic energy causes the dispersion)

Question 3

∆Hsoln =

Answer 3

∆Hsoln = ∆Hsolute + ∆Hsolvent + ∆Hmix

Question 4

Explain why gases generally have a negative ∆Hsoln when dissolved in a liquid?

Answer 4

Energy is required

Question 5

∆Hmix = ?

Answer 5

∆Hmix = | ∆Hsolute + ∆Hsolvent |

Question 6

Polar solutes dissolve in ?

Answer 6

Polar solutes dissolve in POLAR SOLVENTS

Question 7

Do nonpolar solvents dissolve in polar solvents

Answer 7

NO and opposite is true

Question 8

Would you expect this solution to form?
Carbon tetrachloride (CCl4) and hexane (C6H14)

Answer 8

Yes both are non-polar (only LDF forces broken and formed

Question 9

Would you expect this solution to form?
Propane (C3H8) and water (H2O)

Answer 9

No H2O interactions are much stronger than LDF that would form with propane

Question 10

Would you expect this solution to form?
Sodium chloride (NaCl) and hexane (C6H14)

Answer 10

No ion-ion interactions in NaCl are much stronger than the ion-induced dipole that could form hexane

Question 11

Would you expect this solution to form?
Calcium chloride (CaCl2) and water (H2O)

Answer 11

yes, the strength of H2O interactions are similar to ion-dipole interactions formed with CaCl2

Question 12

Would you expect this solution to form?
Water (H2O) and methyl alcohol (CH3OH)

Answer 12

Yes, the strength of H2O interactions are similar to the dipole-dipole interactions and h-bonds formed with methyl alcohol

Question 13

Saturated solution - rate of dissolution =?

Answer 13

Rate dissolution = Rate precipitation

Question 14

Do unsaturated solutions have an equilibrium

Answer 14

no equilibrium

Question 15

Super saturated solutions

Answer 15

instable solution crystallization?

Question 16

If ∆Hsoln > 0

Answer 16

Dissolution of salts in water is usually endothermic

Solubility increases with increasing temperature

A saturated solution may become supersaturated if it is cooled (∆Hsoln > 0)

A saturated solution may become unsaturated if it is heated (∆Hsoln > 0

Question 17

If ∆Hsoln < 0

Answer 17

Solubility decreases with increasing temperature

Dissolution of gases in water is exothermic

Question 18

Mole fraction (X)

Answer 18

Xa = moles of A / total moles in solution

Question 19

Calculate X NaCl and X H2O of a solution composed of 36.00 g NaCl dissolved in
150.00 g H2O.

Answer 19

1

Question 20

Calculate the mass of NaCl and the mass of H2O in 200.00 g of a solution, if X NaCl = 0.200 and X H2O = 0.800 .

Answer 20

1. Calculate the mol of NaCl and the mol of H2O in a total of 1 mol of compounds.
   In 1 mol total: mol NaCl + mol H2O = 1 mol
2. Calculate m NaCl and m H2O in this solution.
3. Calculate the total mass of this solution
4. Convert 200.00 g solution to mass NaCl and mass H2O.

1.10 x 10^2g H2O - page 18

Question 21

Molality (m)

Answer 21

molality = mol of solute / Kg of solvent

Question 22

Molality of a solution containing 20.00 g NaCl in 200.00 g H2O.

Answer 22

1.711m NaCl p.19

Question 23

Calculate the mass (in g.) of NaCl and the mass of H2O needed to prepare 100.00 g of a 0.100 m NaCl solution.

Answer 23

1. Calculate the mass of NaCl in 1kg of H2O
2. Calculate the total mass of solution
3. Calculate the mass of NaCl in 100.00g solution
4. Calculate the mass of H2O
   99.42g H2O p.20

Question 24

Molarity (M)

Answer 24

M = mol of solute / L of solution

Question 25

Molarity of a solution made by dissolving 100.00 g NaCl in enough water to make 1.00 L of solution.

Answer 25

1.71 mol/L NaCl p.22

Question 26

Dilution formula

Answer 26

V1C1=V2C2

Question 27

Molarity of a diluted solution obtained by dilution of 10.00 mL of 2.00 M NaCl solution to 200.00 mL with water.

Answer 27

1.00 x 10^-1M NaCl p.24

Question 28

Equivalence point

Answer 28

Point at which equivalent amount of base has been added to acid. (mol H+ = mol OH–) Neutralization is reached. (No more HCl, no excess NaOH)

Question 29

Determination of [HCl] of a HCl solution if a 50.00 mL sample of this solution is titrated with 1.00 x 10-2 M NaOH solution and 15.25 mL of NaOH are needed to reach the equivalence point.

Answer 29

1. Write balanced equation 2. Calculate moles of base used 3. Calculate moles of HCl in sample 4. Calculate [HCl] in sample 3.05 x 10^-3M HCl p.29

Question 30

Mass percent (m%)

Answer 30

Mass % solute = ( mass of solute / total mas of solution ) x100

Question 31

Mass percent of NaCl if 15.00 g NaCl is dissolved in 100.00 g H2O.

Answer 31

13.04% NaCl

Question 32

Conversion of X to m

Answer 32

(Mol solute / 1 mol solution) -> (mol solute / 1 Kg solvent)

Question 33

Calculate the molality of an aqueous NaCl solution if the mole fraction of NaCl is 0.024

Answer 33

1.4 m NaCl P.34

Question 34

Conversion of m % to m:

Answer 34

(g solute / 100g solution) -> (mol solute / 1 Kg solvent)

Question 35

HCl is sold as 36 % aqueous solution. Express it concentration in molality.

Answer 35

15 m HCl p.35

Question 36

Conversion of m % to X:

Answer 36

(g solute / 100g solution) -> (mol solute / mol solution)

Question 37

Convert Molality to molarity

Answer 37

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