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The density of Silver (Ag) is \[10.5gc{m^{ - 3}}\]. How do you find the mass of Ag that occupies \[965c{m^3}\] of space?

Answer
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Hint In order to find the mass of the silver that occupies certain space, we must first have an idea about what a density is. Density is referred to as the mass per unit volume. Density is usually represented by the Greek symbol rho.

Complete step by step solution:
First let us understand about what density is. Density is referred to as the mass per unit volume. Density is usually represented by the Greek symbol rho. The equation for density is given as:
\[\rho = \dfrac{m}{V}\]……. (1)
Where, \[\rho \] is the density of the substance.
m is the mass of the substance.
V is the volume of the substance.
From the above equation, we can say that the density is inversely proportional to the volume. This means that the denser material will be having less volume.
Now let us move onto the given question.
In the question density of silver is given as \[10.5gc{m^{ - 3}}\]
The volume of silver is given as \[965c{m^3}\].
The equation (1) is rearranged as
\[m = \rho \times V\]
\[m = 10.5gc{m^{ - 3}} \times 965c{m^3} = 1.01 \times {10^4}g\]
The mass of the silver is found to be \[1.01 \times {10^4}g\].

Note: We have to remember certain points in order to get a clarity on density. They are:
- The density of the substance will vary with temperature and pressure.
- The density will vary only a very little for the solid and liquid but greater for the gases.
- Density is said to be an intensive property.