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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  the definition of the average density\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

ρ=mV\rho = \dfrac{m}{V}

Where:

  • ρ\rho is the average density of the gold .
  • mm is the mass of the gold .
  • VV is the volume of the gold

Givens\textbf{Givens}: ρ=1.93×104kg/m3\rho = 1.93 \times 10^4 \mathrm{kg/m^3} , m=31.103gm =31.103 \text{g} .

Plugging\textbf{Plugging} known information to get :

ρ=mVV=mρ=(31.103g)(1kg103kg)1.93×104=1.61×106\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*}

[1.61×106m3106cm31cm3=1.61cm3]\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 ]

V=1.61cm3\boxed{ V = 1.61 \mathrm{cm^3} }

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