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Calculation of Mass

  1. Jan 29, 2012 #1

    MacLaddy

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    Gold Member

    1. The problem statement, all variables and given/known data

    The density of Benzene at 15 degrees Celsius is [itex]0.8787 \frac{g}{mL}[/itex] Calculate the mass of 0.1500 L of Benzene at this temperature.

    2. Relevant equations

    [itex]Density=\frac{mass}{volume}[/itex]


    3. The attempt at a solution

    The part of this problem that is throwing me off is the g/mL, and L measurements. What is have done is I took the density and multiplied it by 1000 to get g/L instead of g/mL. The equation then looked like this.

    [itex]878.7\frac{g}{L} = \frac{mass}{0.1500L}\rightarrow
    878.7(0.1500)=mass\rightarrow
    131.805g = mass[/itex]

    I'm not sure if I am correct with that result, but even if I am I believe I am making a mistake with how I distribute L and g/mL across my multiplication. Any advice on this problem would be appreciated.

    Mac
     
    Last edited: Jan 29, 2012
  2. jcsd
  3. Jan 30, 2012 #2

    Borek

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    Staff: Mentor

    You did OK. You could as well convert volume to mL (0.1500L = 150.0mL) - final result would be identical.
     
  4. Jan 31, 2012 #3

    MacLaddy

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    Thanks for the double-check, Borek.

    I realized what was confusing me so badly. I knew that if I multiplied L by 1000 I could get mL, yet I was seeing that I also needed to multiply my g/mL by 1000 to get g/L. I was thinking, "how can I multiply both by 1000, shouldn't I be dividing one of the figures?" Anyhow, it finally dawned on me, it's g/mL, not just mL.

    Thanks again.
    Mac
     
    Last edited: Jan 31, 2012
  5. Jan 31, 2012 #4

    Borek

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    Staff: Mentor

    Quite the opposite - mL has to be divided by 1000 to be converted to L. Using dimensional analysis:

    [tex] 10\ \text{mL} = 10\ \text{mL}\frac{1\ \text{L}}{1000\ \text{mL}} = 0.01\ \text{L}[/tex]
     
  6. Jan 31, 2012 #5

    MacLaddy

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    Gold Member

    You're correct, Borek, and I misspoke. I was actually referring to multiplying L by 1000, as in your post here,

    I've corrected the error in that post.
     
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