Volume of Iron (dealing with the units)

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Homework Help Overview

The discussion revolves around calculating the volume of one mole of iron, given its molar mass and density. The problem involves unit conversions and understanding the relationship between mass, volume, and density.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the correct mass to use in calculations, questioning the original poster's use of units. There is discussion about converting units from kilograms to grams and from cubic meters to cubic centimeters.

Discussion Status

Some participants have provided guidance on unit conversions and clarified the proper mass to use. There is ongoing exploration of the correct approach to obtain the volume, with multiple interpretations of the necessary steps being discussed.

Contextual Notes

There is confusion regarding the conversion of density units and the implications of using different unit systems. Participants are also addressing the need to ensure that all units are consistent throughout the calculations.

blipped
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Homework Statement


What is the volume of one mol of iron where iron is 55.8 g/mol and the density of iron is 7.86 x 10^3 kg/m^3 ?


Homework Equations


p = m/v


The Attempt at a Solution



I can get the numerical answer (checked back of book), but I'm not sure how my units worked out?

v = m/p = (55.8 g/mol) / (7.86 x 10^3 kg/m^3)

I just knock off the 10^3 and divie 55.8 by 7.86 and I get the answer of 7.10 and my units would be cm, but why? How did this work out?

Thanks.
 
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Hi blipped,

blipped said:

Homework Statement


What is the volume of one mol of iron where iron is 55.8 g/mol and the density of iron is 7.86 x 10^3 kg/m^3 ?


Homework Equations


p = m/v


The Attempt at a Solution



I can get the numerical answer (checked back of book), but I'm not sure how my units worked out?

v = m/p = (55.8 g/mol) / (7.86 x 10^3 kg/m^3)

I just knock off the 10^3 and divie 55.8 by 7.86 and I get the answer of 7.10 and my units would be cm, but why? How did this work out?

Thanks.

Your mass can't be 55.8 g/mol, can it? That's not the right units for mass. That's the factor to convert moles of iron to grams of iron. So what is the mass with the right units, and how do you get it?

(And the units you get at the final step is not just centimeters; that would be for a length.)
 
Thanks for help!

I suppose then it's just 55.8 g. The problem stated that there are 55.8 grams or iron per mol, so that's what I put..

I figured out what I'm supposed to do I think.. Just take the cube root of 7.86 x 10^3 kg/m^3 in order to get them to be the same units (ratio of mass to volume) then divide

55.8 g / 7.86 g /m^3 which is 7.1 m^3 for the volume..

The answer is correct according to the book but is that the right way to do it?
 
blipped said:
Thanks for help!

I suppose then it's just 55.8 g. The problem stated that there are 55.8 grams or iron per mol, so that's what I put..
That's right; you convert the 1 mole of iron to grams of iron, and so your expression will not have units of moles in it at all.
I figured out what I'm supposed to do I think.. Just take the cube root of 7.86 x 10^3 kg/m^3 in order to get them to be the same units (ratio of mass to volume) then divide
I'm not sure what you are referring to with the cube root. I don't see where you took a cube root. (I don't see where it is needed, either.)
55.8 g / 7.86 g /m^3 which is 7.1 m^3 for the volume..

I think you went the wrong way in the unit conversion.

<br /> 7860 \mbox{ kg/m}^3 \neq 7.86 \mbox{ g/m}^3<br />

But didn't you want the answer to be in cm^3? So if you convert both kg \to g and m^3\to\mbox{cm}^3, what do you get?
 

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