Ch 11 Solutions

Question 1

A homogeneous mixture of two more substances in any state of matter is a?

Answer 1

Solution

Question 2

The substance present in smaller amounts is the?

Answer 2

Solute

Question 3

The substance present in larger amounts is the?

Answer 3

Solvent

Question 4

__________ contains the maximum amount of a solute that will dissolve in a given solvent at a specific temperature

Answer 4

Saturated solution

Question 5

________ contains less solute than the solvent has the capacity to dissolve at a specific temperature

Answer 5

Unsaturated solution

Question 6

_________ contains more solute than is present in a saturated solution at a specific temperature

Answer 6

Super saturated solution

Question 7

Saturation depends on the amount of?

Answer 7

Solute

Question 8

Like dissolved like

Answer 8

I

Question 9

Concentration of a solution is?

Answer 9

The amount of salute present in a given quantity of solvent or solution 

Question 10

Mole fraction is

Answer 10

Moles of A / total moles of A+B+C

Question 11

% by mass is?

Answer 11

Mass of solute
———————————— X 100%
Mass of solute + Mass of solvent

Mass of solute / total mass solution x100

Question 12

Molarity Concentration with M capital =

Answer 12

moles of solute
M = ———————
Liters of Solution

Question 13

MolALITY (m) =

Answer 13

Moles of solute
m = ———————-
Mass of Solvent (kg)

Question 14

What is the percent by mass of NaCl in a solution made by dissolving 4.6 g of NaCl in 500 g of water?

Answer 14

Solvent 4.6g NaCl
Solution 500g water = .5 liters

      4.6g NaCl ————————— x100 = 0.912%  4.6g + 500g H2O

Question 15

Calculate the MolALITY of 4.6 g NaCl in 500g of water

Answer 15

First grams to moles
4.6g/58.5g NaCl=0.0789 mol NaCl

    0.079 mol NaCl  ——————————————- = 0.16M 4.6+500g *1ml/1g *1L/1000mL

Question 17

Calculate the Mole Fraction of 4.6 g NaCl in 500g of water

Answer 17

Moles of NaCl / total moles of both =mole fraction

0.079 moles of NaCl

H2O 500g/18.0g =27.778 mol H2O

0.0789/ 0.0789+27.78 =28.78 0.00283

Question 18

Calculate the MolALITY of 4.6 g NaCl in 500g of water

Lower case m

Answer 18

m= 0.0787 mol / 0.5kg = 0.157m

Question 19

Does solubility increase or decrease with increasing temperature?

Answer 19

Solubility increases with increasing temperature just like it’s easier to dissolve sugar and hot tea than in iced tea

Question 20

The separation of a mixture of substances into pure components on the basis of their differing solubilities is called

Answer 20

Fractional crystallization

Question 21

Solubility of gases ______ with increasing temperature

Answer 21

Decreases

Question 22

Properties that depend only on the number of solute particles in solution, and not on the nature of the solid particles

Answer 22

Colligative properties

Question 23

Change of boiling point temperature =

Answer 23

K(boiling) * m (molality of solution)

K= boiling constant of given solvent

Question 24

Change in Tf (freezing point temperature) =

Answer 24

A freezing point of the pure solvent minus the freezing point of the solution

Question 25

What is the freezing point of a solution containing 478 g of ethylene glycol antifreeze in 3202 g of water? The molar mass of ethylene glycol is 62.01 g

Answer 25

Moles of solute /mass of solvent = m 478g/62.01g mol EG / 3.202kg water =2.41m H2O freeze Kfm = 1.86c/m * 2.41m =4.483 ℃ 0DC -4.483= -4.48dc