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Chem Question

M

Mr Bob

Golden Member
The volume of seawater on Earth is about 330,000,000 mi3. If seawater is 3.5% sodium chloride by mass and has a density of 1.03 g/mL, what is the approximate mass of sodium chloride, in tons, dissolved in the seawater on Earth?

D = M/V

1.03mL = M/330,000,000mi^3

M = 3399000mL/mi^3

Is this correct so far?
 
Originally posted by: Mr Bob
The volume of seawater on Earth is about 330,000,000 mi3. If seawater is 3.5% sodium chloride by mass and has a density of 1.03 g/mL, what is the approximate mass of sodium chloride, in tons, dissolved in the seawater on Earth?

D = M/V

1.03mL = M/330,000,000mi^3

M = 3399000mL/mi^3

Is this correct so far?
Click to expand...

convert cubic miles to liters or milliters first. (1.4 x 10^21 liters)

Then figure out total mass of seawater, multiply that by 0.035
 
Let's see if this is right...

1 cubic mile = 4.16818183 x 10^15

330,000,000 mi^3 X 4.16818183 X 10^15 mL = 1.375500004 X 10^24 mL

D=1.03g/mL
M=?
V=1.3755000004 X 10^24 mL

1.03g/mL = M g / 1.375500004 X 10^24 mL

M/g = 1.416765004 X 10^24 g

Then
1.416765004 X 10^24 X 0.035 = 4.958677514 X 10^22 g

Grams to tons:
4.958677514 X 10^22 g X 1.10231131 X 10^-6 = 5.466006306 X 10^16 tons

Final answer: 5.466006306 X 10^16 tons
 
I prefer to teach students to "multiply by one (showing units)" when doing such problems.

330,000,000 mi3 * 5280 feet /mi * 5280 feet/mi * 5280 feet/mi [note: at this point, your units are now in cubic feet]

*12 in/ft *12 in/ft *12 in/ft *2.54cm/in *2.54cm/in *2.54cm/in *1ml/cm3 [still continuing; if you're following along, your units are now ml of water)

*1.5g/ml [you're at grams of water at this point] * .035 grams sodium/gram water

 
Note, that when you're writing it out this way, you can cancel the units as you go. Since 12 inches = 1 foot, you can write 12in/ft or 1ft/12in.
Since the numerator = the denominator, the fraction has a value of 1.
Determine which of the two fractions to use by what units you want to cancel out. This won't solve all of your problems, but you'd be amazed at how much science can be solved using unit analysis. (at the high school level)
 
If it was homework I would have said it.

Thanks DrPizza, I thought there was more to the problem than a simple converstion. I didn't realize that you could go from mi3 to mL.
 
Alright then, I guess I just have a few of these non "homework" questions when school starts. It will be completely non homework related just purely coincidental to school starting. For some odd reason my curiosity will increase ten fold so I'll be blasting a few of these. Hope everyone here is so kind as to help me!
 
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