CHM110 Midterm 1

Question 1

What is
1) %w/w
2) %w/v
3) %v/v

Answer 1

1) g solute/100g solution
2) g solute/100mL solution
3) mL of solute/100mL solution

NOTICE that it’s always solute to solution

Question 2

Define molarity and molality and explain temperature dependence

Answer 2

Molarity: mol solute/ L solution (temp dependent
Molality: mol solute/ kg solvent (not temp dependent)

Temp dependent becuase volume of solution changes which temperature, most liquids expand with increases. Molality is dependent on mass though

Question 3

Given %w/w and density, find molairity, molality and mole fraction

Answer 3

1) MOLAIRITY:
- 98%w/w is simply 98g solute in 100g of solution. Multiply by density to cancel out solution
- Units are in g/mL, multiply by 1000 to make g/L
- Divide by molar mass to cancel out the g for a final unit of mol/L
2) MOLALITY
- If 98% solute then 2% is solvent. 98% = 98g and 2% = 2g.
- Multiply molariity by 1000 to make g/kg and then divide by the molar mass of the solute to get mol of solute/mass of solvent
3) MOLE FRACTION
- If 98% solute then 2% is solvent. 98% = 98g and 2% = 2g. Set up as 98/2 and find the moles of each using the molar mass
- Divide desired mole value by total moles (total moles can be greater than 1 but the mole fractions SHOULD equal 1) to find the mole

Question 4

Given molairity and molality, find mole fraction, density and %w/w

Answer 4

1) MOLE FRACTION:
- Use molality since it is in mol mol solute/kg solvent
- Find moles of solvent by converting to g and then using molar mass to convert to mol
- Add moles together to find total moles
- Divide mol of solute by total moles to find mole fraction
2) DENSITY:
- Calcualte mass of solute (moles is the molairity value given multiplied by molar mass of solute)
- Calcualte mass of solvent is moles divided by molality (mol x kg/mol) and turn into g
- Add both masses and assume volume of solution is 1L
- Density = total mass/volume
3) WEIGHT BY WEIGHT
- Calculate mass of solute (use molairity for moles and molar mass)
- Calculate mass of solute (use moles of solute/molality which is mol * kg/mol and convert to g
- Find the total mass of the solution
- Divide desired mass by total mass and multiply by 100

Question 5

Equation for density

Answer 5

Density (g/L) = mass/volume

Question 6

Unit conversion of L to M^3

Answer 6

1L = 0.001 cubic meter
Divide by 1000

Question 7

Given density and mole fraction, find, molairity and molality, %w/w

Answer 7

1) %W?W (FIND THIS ONE FIRST)
- Use mole fraction to find the moles of each reactant. So XA = nA/nT (total moles) and then you can find the mass by multiplying by the molar mass but KEEP IN MIND the nT doesn’t cancel out till the end when you divide both values with the molality
- Then use these moles and calculate the mass of each in g
- Add both these grams together
- Divide solute in grams by total and multiply by 100%

2) & 3) Just follow same method as above

Question 8

Given molality and density, find molairity, mole fraction and %w/w

**even without asking to find %w/w this method is still useful

Answer 8

1) MOLAIRITY
- Calcualte moles of solute using the molality
Moles = molality x mass of solvent (assume 1kg)
= mol/kg x kg and convert to grams
- Calculate mass of solute by multiplying by molar mass
- Combine both mass of solute and mass of solvent (assumed to be 1kg)
- Divide mass of solution by density to obtain the volume
- Molairity is moles of solute/divided by volume
2) %W/W
- Use previous mass numbers obtained
3)

Question 9

What does 1 mol equal

Answer 9

6.022 x 10^23

Question 9

Determine empirical formula from combustion analysis

Answer 9

• Convert masses of compounds to moles of compounds
• Calcualte mole of element in compound (H2 is multiply by 2 for example)
• Convert to mass of element
• Continue as usual

Question 10

Mass percent formula

Answer 10

mass of element (moles x molar mass)/molar mass of compound

Question 11

Why is theoretical yield not obtained?

Answer 11

Side reactions, reaction stopping before complete, physical losses (like solid clinging to another paper)

Question 12

Why does water dissociate ionic compounds into ions

Answer 12

Attraction between O in water attracted to cations and H attracted to anions is strong enough to break attraction between ions

Question 13

Electrolyte and strong, weak, and non

Answer 13

When dissolved in water produced ions and conducts eletricity

Strong - acid/base and ionic
Weak - acid/base
Non - covalent

Question 14

Why is molality greater than molarity usually?

Answer 14

With molairity, adding more solent increases TOTAL volume so smaller fraction

Question 15

Using couloumb’s law, distinguish between bonding and IMF

Answer 15

Bonding forces stronger because larger charges closer together

IMF weaker becuase smaller charges father apart

Question 16

What is LDF

Answer 16

Electron charge is distributed unevenely on atom so creates temporary dipole which induces other temporary dipoles resulting in weak attractions

Question 17

Does weak IMF cause high or low vapour pressure?

Answer 17

High because it is easy for molecules to escape to vapour phase

Question 18

Define polariziability and what kinds of atoms have higher polariziability

Answer 18

How easily electron cloud distorted by other ions or dipoles to cause induced dipole

Heavier, alrger, and negative ions are more polarizable

Question 19

If you need to produce 100mL solution of 4 solutions of 0.250mol/L each and you are provided by four 25mL flasks containing the solutions with 1.00mol/L.

Would pouring the contents into a 100mL beaker provide desired final solution with high precision?

5 points

Answer 19

1. Combining solutions of four 25mL flasks would give approximate final volume of 100mL
2. Calculate mol of final (0.025) of 0.1L so concentration would be 0.25 mol/L
3. Pouring liquid out of volumetric flask is not quantitative transfer (one of those would be using pipette or funnel) and total volume would be less than 100mL
4. Adding additional water to rinse the flasks would give final solution volume greater than 100mL
5. Not appropriate method for producing desired final solution

Question 20

When do moles of solution change?

Answer 20

When pressure or temperature changes, but not with volume or concentration

Question 21

Starting with an equation for molality and mole fraction of solute A in water, express mole fraction as a function of ONLY molality!

Answer 21

1) Find nw
- mass w = nw x MMw
- assume mass of 1kg (1000)
- should be around 55.5 mol
2) Find na
- molality = nA/massw
na = molality x mass (1)
na = m
3) Combine to mole fraction equation
- molality / molality + 55.6

Question 21

Mixture of two salts (mass given) in excess other salt creates a percipitatte of a given mass. Calculate moles of each

Answer 21

1) Find moles of percipitate
2) Set x + y = answer from 1 and rearrange to isolate y
REMEMEBER when to double if 2 moles makes 1 mol
3) mass salt A + mass salt B = mass of two salts

Mass = (x and molar mass)

Rearrange and solve for x

Question 22

What is the equation for molar volume of a gas

Answer 22

MV = volume/moles

Question 23

Two different gases, which has higher average kinetic energy

Answer 23

Temperature is a measure of average kinetic energy so they would be the same

Temperature is proportional to average kinetic energy

Question 24

Avogardro’s principle

Answer 24

Equal volumes of gases at the same temperature and pressure will contain equal number of molecules (even when they are different gases!)

Question 25

equation for partial pressur eof a gas A

Answer 25

pA = mole fraction A times total pressure

Question 26

Deviations from real gases

Answer 26

Low temperature and high pressure
- Gases do indeed have volume
- Attractive and repulsive forces exist among gas particles (because atoms exist as charged subatomic aprticles and many bonds are polar

Question 27

Volume required for neutralization?

Answer 27

Normality 1 x volume 1 = normality 2 x volume 2

Normality: equivalents bsaed on how many hydrogen it donates, multiply molairity by this value

Then isolate and solve like c1v1 = c2v2 problem