Page 2 of 4

### Re: Density Lab B

Posted: February 3rd, 2020, 6:53 pm
Someone pls answer i'm booooooored

### Re: Density Lab B

Posted: February 14th, 2020, 4:04 pm
```[1) Gases are composed of many tiny particles traveling in random straight-line motion.

2) The volume of the gas particles is negligible compared to the total volume of the container.

3) The attractive and repulsive forces between particles are negligible.

4) Collisions are completely elastic.

5) The average K.E. is proportional to the absolute temperature of the gas.```

### Re: Density Lab B

Posted: February 15th, 2020, 9:27 am
All correct! Your Turn!

### Re: Density Lab B

Posted: February 15th, 2020, 4:22 pm
Ok,

1) What is the concentration in ppm of 5.5 milligrams of salt dissolved in 300 mL of water?

2) An object is placed in a graduated cylinder filled with 500 mL of water. The water level rises to 550 mL. Assuming the object is completely submerged, what is the density of this object in kg/m^3 if the object weighs 1.916 N?

3) There are 2.9 moles of oxygen gas held in container that has a volume of 100 mL. What is the number density of the oxygen gas in molecules/m^3?

### Re: Density Lab B

Posted: February 21st, 2020, 6:16 pm
1) 8.5 ppm using a density of 2.16 g/cm^3 for salt.
2) 3,910 kg/m^3
3) 1.7*10^27 molecules/m^3

### Re: Density Lab B

Posted: February 23rd, 2020, 10:46 am
Sapphire wrote:
February 21st, 2020, 6:16 pm
1) 8.5 ppm using a density of 2.16 h/cm^3 for salt.
2) 3,910 kg/m^3
3) 1.7*10^27 molecules/m^3
1) I believe the answer is 18.33 ppm.

Explanation: First, we have to convert both of the masses to the same unit (e.g. both have grams for units). 5.5 milligrams is equivalent to 0.0055 grams. 300 mL of water has a mass of 300 grams.
Now, to find ppm: 0.0055 grams / 300 grams x 1,000,000, bringing our final answer to about 18.33 ppm.

**I didn’t use significant figures here because the answer would be a bit inaccurate, as the answer would have 20 ppm (1 sig fig in 300)

2) Correct.

3) I believe the answer is 3.49 x 10^28 molecules/m^3.

Explanation: Since there are two molecules in oxygen gas, we multiply 2.9 moles by two. This brings the number of moles to 5.8, which we can then multiply by Avagadro’s Number (6.022 x 10^23) to get the number of molecules, which is 3.493 x 10^24. To convert mL to m^3, we have to divide by 10^6, so 100 mL / 10^6 = 0.000100 m^3. To find molecules/m^3, we simply divide: 3.493 x 10^24 / 0.000100, bringing our final answer to 3.49 x 10^28 molecules/m^3. (3 significant figures here)

### Re: Density Lab B

Posted: February 23rd, 2020, 11:29 am
azboy1910 wrote:
February 23rd, 2020, 10:46 am
1) I believe the answer is 18.33 ppm.

Explanation: First, we have to convert both of the masses to the same unit (e.g. both have grams for units). 5.5 milligrams is equivalent to 0.0055 grams. 300 mL of water has a mass of 300 grams.
Now, to find ppm: 0.0055 grams / 300 grams x 1,000,000, bringing our final answer to about 18.33 ppm.

**I didn’t use significant figures here because the answer would be a bit inaccurate, as the answer would have 20 ppm (1 sig fig in 300)

2) Correct.

3) I believe the answer is 3.49 x 10^28 molecules/m^3.

Explanation: Since there are two molecules in oxygen gas, we multiply 2.9 moles by two. This brings the number of moles to 5.8, which we can then multiply by Avagadro’s Number (6.022 x 10^23) to get the number of molecules, which is 3.493 x 10^24. To convert mL to m^3, we have to divide by 10^6, so 100 mL / 10^6 = 0.000100 m^3. To find molecules/m^3, we simply divide: 3.493 x 10^24 / 0.000100, bringing our final answer to 3.49 x 10^28 molecules/m^3. (3 significant figures here)
1) Oh, I see. I did volume/volume instead of mass/mass. How should I know when to divide what by what? Are there different types of concentration or is it always the same?

3) I forgot that oxygen gas and oxygen aren't the same thing, and also made a mistake converting the container to m^3. Oops!

### Re: Density Lab B

Posted: February 23rd, 2020, 2:21 pm
Sapphire wrote:
February 23rd, 2020, 11:29 am
azboy1910 wrote:
February 23rd, 2020, 10:46 am
1) I believe the answer is 18.33 ppm.

Explanation: First, we have to convert both of the masses to the same unit (e.g. both have grams for units). 5.5 milligrams is equivalent to 0.0055 grams. 300 mL of water has a mass of 300 grams.
Now, to find ppm: 0.0055 grams / 300 grams x 1,000,000, bringing our final answer to about 18.33 ppm.

**I didn’t use significant figures here because the answer would be a bit inaccurate, as the answer would have 20 ppm (1 sig fig in 300)

2) Correct.

3) I believe the answer is 3.49 x 10^28 molecules/m^3.

Explanation: Since there are two molecules in oxygen gas, we multiply 2.9 moles by two. This brings the number of moles to 5.8, which we can then multiply by Avagadro’s Number (6.022 x 10^23) to get the number of molecules, which is 3.493 x 10^24. To convert mL to m^3, we have to divide by 10^6, so 100 mL / 10^6 = 0.000100 m^3. To find molecules/m^3, we simply divide: 3.493 x 10^24 / 0.000100, bringing our final answer to 3.49 x 10^28 molecules/m^3. (3 significant figures here)
1) Oh, I see. I did volume/volume instead of mass/mass. How should I know when to divide what by what? Are there different types of concentration or is it always the same?

3) I forgot that oxygen gas and oxygen aren't the same thing, and also made a mistake converting the container to m^3. Oops!
1) When you are finding ppm, it doesn’t have to necessarily be mass/mass. It could also be volume/volume, or even moles/moles, but they have to be the same unit (e.g. grams/grams and not grams/kilograms). There are also some cases where you can mass/volume, but I would suggest doing mass/mass, moles/moles, or whatever you prefer. For the dividing part, I simply divide mass of solute/mass of solvent, especially if there’s a small amount of solute, but don’t be misled by this because there could be a large amount of solute. In this case, I would instead do mass of solute/mass of solution.

### Re: Density Lab B

Posted: February 24th, 2020, 3:12 pm
Okay I'll post a few:

1) How is molality affected by temperature?

2) A sample of gas occupies a volume of 11.1 L at 1.23 atm and 0.0 degrees Celsius. What volume does the sample of gas occupy at 3.15 atm and 100.0 degrees Celsius?

3)All of the below are types of densities except:
a) Mass density
b) Number density
c) Area density
d Volume density

### Re: Density Lab B

Posted: February 25th, 2020, 7:11 am
NewSciolyer wrote:
February 24th, 2020, 3:12 pm
Okay I'll post a few:

1) How is molality affected by temperature?

2) A sample of gas occupies a volume of 11.1 L at 1.23 atm and 0.0 degrees Celsius. What volume does the sample of gas occupy at 3.15 atm and 100.0 degrees Celsius?

3)All of the below are types of densities except:
a) Mass density
b) Number density
c) Area density
d Volume density
1) Molality is not affected by temperature
2) 5.92 L
3) d-Volume density