Convert 100 grams of a solution to milliliters with a density of $681.9\rm~\frac{kg}{m3}$. It also has a density of 5.69 pounds per gallon.

Are any or all of these correct?

$$\rm100~g \times \frac{1~mL}{681.9 \frac{g}{cm^3}} = 0.146649~mL\quad(\approx 0.15~mL)$$

$$\rm1,000~g \times \frac{1,000~mL}{681.9~\frac{g}{cm^3}} = 1,466.4906~mL\quad(\approx 1,500~mL)$$

$$\rm1~kg \times \frac{1~L}{681.9~\frac{kg}{m^3}} = 1.4664906~L (\approx1.5~L)$$

Or, can it be as simple as: 100 g x 681.9 = 68.190 mL? The unit kg/m3 is what's confusing me. I'm used to g/cm3 or g/mL for densities. Does 681.9 g/cm3 = 681.9 kg/m3, while scaled-up a thousand-fold? Can I consider using the same density with the math involved by substituting kg/m3 for g/cm3 in the equation, or do I have to utilize a conversion within the metric system?

  • $\begingroup$ user26967, the original post was off-topic primarily because you didn't show any of your efforts toward solving/understanding the problem. Now that you've edited in your guesses, there's a good chance it will be re-opened. $\endgroup$ – hBy2Py Feb 19 '16 at 15:55
  • $\begingroup$ Thank you for the complete understanding. I greatly appreciate the time you invested in answering my question. Thanks again! $\endgroup$ – user26967 Feb 21 '16 at 21:25

Concerning the units in your equation none of them is correct (As all your formulations would end up in a square volume unit).

Definition of volumetric mass density

The density $ \rho $ is given as

$$ \rho = \frac{m}{V} $$

resulting in a mass and density dependent Volume as

$$ V = \frac{m}{\rho} $$

Direct way

Although you could calculate this by just substituting the physical quantities by the given values and units, and then take care of the unit conversion afterward:

$$ V = \frac{m}{\rho} = \frac{100\,\mathrm{g}}{681.9\,\mathrm{kg}\,\mathrm{m^{-3}}} =\\= \frac{100\,\mathrm{g}\,\mathrm{m^{3}}}{681.9\,\mathrm{kg}} = \frac{100\,\mathrm{g} \cdot \left(10^2\,\mathrm{cm}\right)^{3}}{681.9 \cdot (10^3\,\mathrm{g})} =\\= \frac{100 \cdot 10^6\,\mathrm{cm}^{3}}{681.9 \cdot 10^3} = \frac{100 \cdot 10^3}{681.9}\,\mathrm{cm}^{3} =\\\approx 146.65\,\mathrm{cm}^{3} \equiv 146.65\,\mathrm{ml} $$

I would suggest to convert the density from $ \mathrm{kg}\,\mathrm{m^{-3}} $ to $ \mathrm{g}\,\mathrm{cm^{-3}} $ as a first step:

Easier way

Of course easiness is something subjective, but I believe this is easier to use.

Calculate conversion-factor for density

First the combined conversion-factor for the units-fraction is calculated

$$ \frac{\mathrm{kg}}{\mathrm{m}^3} \stackrel{\left[1\right]}{=} \frac{\left(10^3\,\mathrm{g}\right)}{\left(10^2\,\mathrm{cm}\right)^3} \stackrel{\left[2\right]}{=} \frac{10^3\,\mathrm{g}}{10^6\,\mathrm{cm}^3} \stackrel{\left[3\right]}{=} 10^{-3}\,\frac{\mathrm{g}}{\mathrm{cm}^3} $$

  1. Start with substituting each units SI prefix:
  2. Care has to be taken with the conversion of $ \mathrm{m}^3 $, since the power applies to the whole substituted expression ($ \mathrm{m}^3 = \left(10^2\,\mathrm{cm}\right)^3 $).
  3. In the last step the exponential expressions are summarized ($ \frac{10^3}{10^6} = 10^{3-6} = 10^{-3} $).

Then calculate the new numeric value of the density in the target unit:

$$ \rho = 681.9\,\mathrm{kg}\,\mathrm{m^{-3}} \stackrel{\left[4\right]}{\equiv} 681.9 \cdot \left(10^{-3}\,\mathrm{g}\,\mathrm{cm}^{-3}\right) \stackrel{\left[5\right]}{=} 0.6819\,\mathrm{g}\,\mathrm{cm}^{-3} $$

  1. Substitute the unit with the previously calculated expression
  2. Multiply the numeric terms

Calculate the volume

Now the unit-adjusted density is used:

$$ V = \frac{m}{\rho} = \frac{100\,\mathrm{g}}{0.6819\,\mathrm{g}\,\mathrm{cm^{-3}}} = \frac{100}{0.6819}\,\mathrm{cm^{3}} \approx 146.65\,\mathrm{cm}^{3} \equiv 146.65\,\mathrm{ml} $$


According to SI Prefixes the following applies:

$ 1\,\mathrm{m} \equiv 10^2\,\mathrm{cm} \\ 1\,\mathrm{kg} \equiv 10^3\,\mathrm{g} $

And of course:

$ 1\,\mathrm{cm}^3 \equiv 1\,\mathrm{ml} $

| improve this answer | |
  • $\begingroup$ Just read this and saw, that I violated the rule, not to give a full answer right away. Anyway I find it hard for this question to give a not full answer and still answer the question helpfully. Should I edit the question to remove the full solution, or delete it? Or can it stay, because I at least gave an explanation how to get there? $\endgroup$ – lcnittl Feb 21 '16 at 19:44
  • 2
    $\begingroup$ I do not think that it would be helpful to remove anything from your answer. $\endgroup$ – aventurin Feb 21 '16 at 20:17
  • $\begingroup$ Thank you for the complete understanding. I greatly appreciate the time you invested in answering my question. Thanks again! $\endgroup$ – user26967 Feb 21 '16 at 21:26

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