Blog: The liquid CHBr3 has a density of 2.89 g dm-³. What volume of this liquid should be measured to contain a total of 4.8×10²⁴ molecules of CHBr3 ?

Hello!

The molar mass of $\displaystyle{C}{H}{B}{r}_{{3}}$ (bromoform) is about

$\displaystyle{12}+{1}+{3}\cdot{80}={253}{\left(\frac{{g}}{{{m}{o}{l}}}\right)}.$

Therefore one mole of this substance has a mass of about $\displaystyle{253}{g{.}}$

One mole of any substance contains $\displaystyle{N}_{{A}}\approx{6}\cdot{{10}}^{{{23}}}$ molecules (this constant is called Avogadro's constant). Hence the given number of molecules represents $\displaystyle\frac{{{4.8}\cdot{{10}}^{{{24}}}}}{{{6}\cdot{{10}}^{{{23}}}}}={8}$ (moles), and from the above paragraph their mass is about $\displaystyle{8}\cdot{253}={2024}{\left({g}\right)}.$

Finally, volume may be computed as mass divided...

Hello!

The molar mass of $\displaystyle{C}{H}{B}{r}_{{3}}$ (bromoform) is about

$\displaystyle{12}+{1}+{3}\cdot{80}={253}{\left(\frac{{g}}{{{m}{o}{l}}}\right)}.$

Therefore one mole of this substance has a mass of about $\displaystyle{253}{g{.}}$

Finally, volume may be computed as mass divided by the density, because $\displaystyle\rho=\frac{{m}}{{V}}.$ In this case it is

$\displaystyle{V}=\frac{{{2024}{g}}}{{{2.89}\frac{{g}}{{{{\left({c}{m}\right)}}^{{3}}}}}}\approx{700}{{\left({c}{m}\right)}}^{{3}}.$

This is the same as 0.7 $\displaystyle{d}{{m}}^{{3}}.$

Note that the density of $\displaystyle{C}{H}{B}{r}_{{3}}$ is incorrectly stated in the question as $\displaystyle{2.89}\frac{{g}}{{{{\left({d}{m}\right)}}^{{3}}}}.$ Actually it is $\displaystyle{2.89}\frac{{g}}{{{{\left({c}{m}\right)}}^{{3}}}}.$ This liquid is much more dense than water (about $\displaystyle{1}\frac{{g}}{{{c}{{m}}^{{3}}}}$ ).