Question

Pure gold has a density of $ 19.32g/c{{m}^{3}} $. How large would a piece of Gold be if it had a mass of $ 318.97g $ ?

Answer 1

Hint: Density is defined as the mass of a substance divided by its volume, it acts as a parameter for recognizing how a substance is distributed in space. Gold is represented by the chemical symbol $ \left( Au \right) $ and has a molar mass of $ 196.96657amu $ . The word Large in the question represents volume as volume is the parameter missing here.

Complete step by step answer:
Gold is also known as one of the inert elements out there because of its high tendency to not react and stay inert under any circumstances. However apart from being an inert or noble metal it also is quite lustrous and rare to find which has made it quite an expensive commodity around which the economies of various human societies revolve.
We are provided with the density of pure gold alongside the mass of the bulk piece whose size requires to be determined, for such a determination the point which should be kept in mind is that when mass is divided by density we obtain volume and this volume is what is the required size asked in the question. To do so let us first write the formula for density:
 \[Density=\dfrac{Mass}{Volume}\]
So, $ Volume=\dfrac{Mass}{Density} $
Thus, volume is $ \dfrac{318.97}{19.32}c{{m}^{3}}=16.50983437c{{m}^{3}}\approx 16.51c{{m}^{3}} $
Now, $ 1l=1000c{{m}^{3}} $ so the required solution is $ 0.01651l $ .
It is only soluble in Aqua regia which is a $ 3:1 $ mixture of $ HCl\And HN{{O}_{3}} $ and is used in electronic chips extensively due to its high conductivity and inert nature.

Note: While calculating the volume as in above case you must be very careful about the unit of various entities provided and must ensure necessary conversions to bring all the entities used in the formula into similar units so that any error does not creep in because of units.