Nancy answers:

20 = kf x m
20 = 1.86 x m
m = 10.8 = moles glycol / Kg water
Denity water = 1.0 g/mL
6.50 L=> 6.50 Kg
Moles glycol = 10.8 x 6.50 = 70.2
Mass ethylene glycol = 70.2 mol x 62.068 g/mol = 4357 g
V = mass / density = 4357 / 1.11 = 3925 mL => 3.925 L

Delta T = 0.512 x m = 0.512 x 10.8 = 5.53 °C
boiling point = 105.53 °C

Nancy answers:

I’m not going to do your homework for you.

But I will help you get closer.

For number three:
what is the difference between the two systems? How do you convert one temperature to the other system?

For number two –if you wanted to see how much you and a dog weigh together, yo weigh yourself first, then hold the dog and step on the scale. Same idea here.

For number one I don’t know. I think it may be a little bit of a trick question.

Nancy answers:

1. A sample of an unknown metal has a mass of 35.4 g and a volume of 3.11cm^3.
Calculate its density. P= blank g/cm^3
density=35.4g/3.11cm³
=11.254cm/cm³

2. Glycerol is a syrupy liquid often used in cosmetics and soaps. A 2.50-L sample of pure glycerol has a mass of 3.15 * 10^3g.
What is the density of glycerol in grams per cubic centimeter? P= blank g/cm^3
density=3,150g/2,500cm³ 1L=1000 cm³
= 1.26g/cm³

3. Ethylene glycol (antifreeze) has a density of 1.11 g/cm^3 .
What is the mass in grams of 388 mL of this liquid?
What is the volume in liters of 3.44 kg of this liquid?
Mass in grams of 388 mL=388 cm³(1.11g/cm³) 1mL=1cm³
=430.68 g
volume in liters of 3.44 kg=[3,440 g/(1.11g/cm³)](1 L/1000cm³)
=3.099L

4. A block of metal has a volume of 13.2 in^3 and weighs 5.48 lb .
What is its density in grams per cubic centimeter?
Density=[5.48 lb(454g/1 lb)/13.2 in³](1 in/2.54cm)³
=11.502 g/cm³

5. A typical backyard swimming pool holds 130 yd^3 of water.
What is the mass in pounds of the water?
130yd³ water=130yd³(0.9144m/1 yd)³(2200 lb/m³)
=218,662.689 lb

6. A backpacker carries 3.0 L of white gas as fuel for her stove.
How many pounds does the fuel add to her load? Assume the density of white gas to be 0.79 g/cm^3

3L=3L(0.79 g/cm³)(1000 cm³/1L)(1 lb/454 g)
=5.22 lb

Nancy answers:

The equation that you need to use for this problem is:

Delta T= Kf * i * m

In this situation your i value is just going to be equal to 1 because ethylene glycol doesn’t dissociate in water. So, you know your Delta T value is equal to 26.0 degrees Celsius because the normal freezing point of water is 0 degrees Celsius. You also know the value of Kf so you need to set the equation up so you can solve for m.

M= T/(Kf * i)
m = 26 degrees C /(1.86 CKg*1) = 13.98

Next, molality (m) is equal to moles of solute per Kg of solvent. So,

13.98 1/kg = moles ethylene glycol / 1.00 kg H2O (1 L H2O = 1 kg H2O)

Multiply each side by 1.00 kg H2O

13.98 = moles ethylene glycol

-> moles ethelene glycol = grams ethylene glycol / molar mass ethylene glycol= g/(g/mol)

13.98 moles = grams ethylene glycol/ (62.01 g/mole ethylene glycol)

Now multiply 62.01 times each side. The moles cancel out and you are left with grams of ethylene glycol.

Grams ethylene glycol = 867 g

—————————————————————————-

Now because you have the grams of ethylene glycol you can solve for the molality and find the boiling point.

Delta T= Kb * i * m

m = moles solute/kg solvent= 13.98 moles ethylene glycol/1.00 kg H2O = 13.98 molal

Delta T = 0.51 C*kg * 1 * 13.98 moles/kg
Delta T= 7.13 degrees C

100 + 7.13 = 107.13 degrees C would be the boiling point

Hope this helps

Nancy answers:

1) there is a total of 1 mole of gas, in 20 dm3 (=0.02 m3) at a temp of 400K
so you need to use PV = nRT with R (the gas constant) = 8.31

The methane will account for 20% of that total pressure and the oxygen for the remaining 80%, since pressure is independent of the type of gas particles that is causing it, only the number of particles, and the temperature they’re at.

Once the methane has all burned, Let’s check the equation:
CH4 + 2O2 ———> CO2 + 2H2O
That means there’ll be 0.2 mol of carbon dioxide, 0.4 mol of steam, and 0.4 mol of unreacted oxygen left over, in excess.
Exactly the same total moles of gas that you had before.
So exactly the same pressure

As for question 2; potassium chloride would start to crystallize out of solution
45.5 – 31 = 12.5
so 12.5 grams of KCl crystals would form.

Part 3;
each kg of water lowers its freeezing point by 1.86 degrees for each mole of substance added.
5/1.86 = 2.69 so 1 kg would need 2.69 moles to lower the fp by 5 degrees
so 10L (10kg) would need 26.9 moles
and 26.9 moles = 1667 grams, which is 1.515 litres
Hope the radiator will hold that much, on top of the 10 litres of water already in there.