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Question

# Gold is sold by the troy ounce (31.103 g). What is the volume of 1 troy ounce of pure gold?

Solution

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From $\textbf{ the definition of the average density}$ we know that :Density The average density of a substance or object is defined as its mass per unit volume

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

Where:

• $\rho$ is the average density of the gold .
• $m$ is the mass of the gold .
• $V$ is the volume of the gold

$\textbf{Givens}$: $\rho = 1.93 \times 10^4 \mathrm{kg/m^3}$ , $m =31.103 \text{g}$ .

$\textbf{Plugging}$ known information to get :

\begin{align*} \rho& = \dfrac{m}{V} \\ V&= \dfrac{m}{\rho} \\ &= \dfrac{\left( 31.103 \text{g} \right) \cdot \left( \dfrac{1 \text{kg}}{10^3 \text{kg}} \right) }{ 1.93 \times 10^4} \\ &= 1.61 \times 10^{-6} \end{align*}

$\left[ 1.61 \times 10^{-6} \mathrm{m^3} \dfrac{10^6 \mathrm{cm^3}}{1 \mathrm{cm^3 }} = 1.61 \mathrm{cm^3} \right ]$

$\boxed{ V = 1.61 \mathrm{cm^3} }$

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