Question

A nugget of gold and quartz weight 100gm. Specific gravity of gold, quartz and nugget is 19.3, 2.6, and 6.4 respectively. Calculate the weight of gold in the nugget.

Answer 1

Hint Specific gravity is the relative density of a material in respect to water. It could be defined as the ratio of the specific weight of the given fluid to the specific weight of pure water at ${{4}^{0}}C$ and is given by the formula,${{d}_{n}}=\dfrac{density(subs\tan ce)}{density(water)}$

Complete Step by step solution:
In the lower classes we have dealt with calculations of specific gravity and density and let us see it in detail.
Specific gravity is the relative density of a material in respect to water. It could be defined as the ratio of the specific weight of the given fluid to the specific weight of pure water at ${{4}^{0}}C$ and is given by the formula, ${{d}_{n}}=\dfrac{density(subs\tan ce)}{density(water)}$
Specific gravity is a unit less and dimensionless positive scalar quantity, and being a dimensionless and unit less quantity specific gravity of a substance is the same in the $S.I.$ and $CGS$ system.
The specific density of a liquid is numerically equal to the relative density of that liquid and substance and for calculation purposes they are used interchangeably.
To calculate the weight of gold by applying the equation (1)
Given that –
Mass of nugget (mn) =100gm
Specific gravity of gold ${{d}_{g}}$= 19.3
Specific gravity of quartz ${{d}_{q}}$ = 2.6
Specific gravity of nugget ${{d}_{n}}$ = 6.4
Now, let Mass of quartz = x
Mass of gold = 100 - x
\[{{d}_{n}}=\dfrac{m}{V}\] ………….(i)
where, m is the mass of nugget and V is the volume of nugget.
Also,$V={{V}_{g}}+{{V}_{q}}$
Hence,$V=\dfrac{{{m}_{g}}}{{{d}_{g}}}+\dfrac{{{m}_{q}}}{{{d}_{q}}}=\dfrac{100-x}{19.3}+\dfrac{x}{2.6}$ …………….(ii)
By substituting equation (ii) in equation (i), we have
\[{{d}_{n}}=\dfrac{m}{\dfrac{100-x}{19.3}+\dfrac{x}{2.6}}\]
\[\Rightarrow 6.4=\dfrac{100\times 19.3\times 2.6}{2.6(100-x)+19.3x}\]
Therefore, by simplifying the above equation, the value of x will be,
\[x=\dfrac{5018-1664}{106.88}\simeq 31.8\]
Thus, mass of gold will be = 100 - x$\Rightarrow m=100-31.8=68.62g$

Additional information:
Relative density is always measured in respect to a reference point. For example, to measure relative density of liquid water is used. Dry air is used to measure the relative density of gases. Density is the ratio of mass and its volume. At an atomic level the density of solid is very high. It is a constant physical quantity which does not depend on temperature and pressure.

Note: Relative density is a dimensionless quantity. Molecular weight is used to determine the specific density if density of air and density gas are evacuated at the same temperature and pressure.