Test 3 Ch 5,6,7

Question 1

If the question GIVES YOU : Pressure and Volume
and ASKS: What is the NEW Pressure or Volume
WHAT law do you use?

P ₁ V ₁ = P ₂ V ₂

Answer 1

Boyle’s Law

P ₁ V ₁ = P ₂ V ₂

Question 2

If the question GIVES YOU : Volume and Temperature
and ASKS: What is the NEW Volume or TEMP
WHAT law do you use?

V ₂ / T ₂ = V ₁ / T ₁

Answer 2

Charles’s Law

V ₂ / T ₂ = V ₁ / T ₁

Question 3

What are the variables in gas law?

Answer 3

T = Temp
V = Volume
P = Pressure
n = # of moles

Question 4

What is used when you are GIVEN: Pressure and Temperature
and ASKS: What is the NEW Temperature or Pressure
What is the law?
P ₂ / T ₂ = P ₁ / T ₁

Answer 4

Guy-Lussac’s Law

P ₂ / T ₂ = P ₁ / T ₁

Question 5

What is used when you are GIVEN: Pressure, Volume and Temperature
and ASKS: What is the NEW Pressure, Volume or Temperature

P ₂V ₂ / T ₂ = P ₁V ₁ / T ₁

Answer 5

Combined gas law

P ₂V ₂ / T ₂ = P ₁V ₁ / T ₁

Question 6

What is used when you are GIVEN: Pressure, Volume, n #of moles and Temperature
and ASKS: What is the NEW Pressure, Volume,n #of moles or Temperature
PV=nRT

Answer 6

Ideal gas law

PV=nRT

Question 7

How would we Find Molar Mass with Ideal Gas law when the given density of the gas is in g/L?
Solve: The density of CO 2at 25.0 ℃ and 1.00 atm is 1.80 g/L.
Calculate its molar mass.

Answer 7

interpret the DENSITY given as g/L as g/V and plug it into
M = (g/V)RT/P
M = (1.8g/1.0L) 0.0821(298) / 1.00atm = 44.03844 = 44.0g/mol

Question 8

PERCENT composition: make 75.0 g of 7.2 % w/w aqueous solution of glucose. How much glucose is required how much water is required? What is the difference if it were v/v (volume mL/ volume mL) and what would you use to measure the volume of solution after you found the solvent?
Hint w/w

Answer 8

7.20% w/w aqueous glucose means 7.20 grams of glucose for every 100 grams of solution.
75 g of solution (7.2 g of glucose / 100 g of solution) = 5.40 g of glucose
75.0 g of solution - 5.40 g of glucose = 69.6 grams of water
You can’t subtract from entire solution to obtain the total volume of water in aqueous solution. You would use volumetric flask or volumetric pipet.

Question 9

Percent concentration w/v: Calculate the number of grams of glucose contained in 45 mL of a 6.0% w/v aqueous glucose solution.

Answer 9

•6.0% w/v means 6.0 g glucose are dissolved in 100 mL solution (not solvent).

45 mL (6.0 g / 100 mL of water) = 2.7 grams of glucose

Question 10

Molarity: How many moles of solute are contained in 1.75 L of a 0.950 M solution?

What if it asked for liters?

Answer 10

Change below equation to solve for moles
moles of solute
M = ———————-
Liter of solution

mol = M x L
1.75 x .95
1.6625 = 1.66 mol
moles of solute
Liter of solution = ———————-
Molarity

Question 11

How would you prepare 1.00 L of a 0.650 M aqueous solution of CaCl2, and How many grams of CaCl2
would be required?

Answer 11

𝑚𝑜𝑙
𝑀 = ———
𝐿
mol = M x L = 0.650 M x 1.00 L = 0.650 mol
Molar mass 111𝑔𝑜𝑓𝐶𝑎𝐶𝑙 2
111𝑔𝑜𝑓𝐶𝑎𝐶𝑙 2
0.650𝑚𝑜𝑙 x———————- = 72.2 gCaCl2
1𝑚𝑜𝑙 𝑜𝑓𝐶𝑎𝐶𝑙2

Weigh out 72.2 g of CaCl 2, and dissolve it in sufficient water to make 1.00 L of solution.

Question 12

What is the final volume of a 0.0015 M solution prepared by dilution of 100 mL of 0.090 M solution?
What unit must the answer be in?

Answer 12

Change the dilution equation to solve for V _f
𝑉𝑖×𝑀𝑖 = 𝑉𝑓×𝑀𝑓
V _i x M _i
V _f = ————————————————-
M _f

100𝑚𝐿 x 0.090 M / 0.0015 M = 6000 mL = 6.0 L
Must be in Liters

Question 13

How do you Calculate the percent by weight of solutions? When given the weight of the solute and solvent.

Answer 13

Add the solute plus the solvent to get the total weight. Then divide the solute by the total weight and multiply by 100 to get the percent w/w