Question

Gold is sold by the troy ounce $\mathbf{\text{(31.103 g)}}$. What is the volume oftroyounce of pure gold?

Short answer

The volume oftroy ounce of pure gold is obtained as:${\text{1.6 cm}}^{\text{3}}$.

Step-by-step solution

Conceptual Introduction

Fluid statics, often known as hydrostatics, is a branch of fluid mechanics that investigates the state of balance of a floating and submerged body, as well as the pressure in a fluid, or imposed by a fluid, on an immersed body.

Given Data 

The density of the substance is the ratio of the mass to that of the volume of the substance.

The density of the Gold is normally considered as$\rho ={\text{19.32 g/cm}}^{\text{3}}$.

Here the mass is given in troy ounce

The mass in gram will be $m=\text{31.103 g}$.

Calculating the volume of the given Substance

As we known that the density is the rate of mass per volume let’s put the given data in the formula given below.

$$\begin{gathered} \rho =\frac{m}{V} \\ V=\frac{m}{\rho } \\ V=\frac{31.103 \text{g}}{19.32 {\text{g/cm}}^3} \\ V=1.6 {\text{cm}}^3 \end{gathered}$$

Therefore, volume of troy ounce gold is ${\text{1.6 cm}}^{\text{3}}$.