Density of Lithium Metal: Calculate with D=m/V

[Faiien]

Homework Statement

What is the density of lithium metal in g/cm^3 if a cylindrical wire with a diameter of 2.40mm and a length of 15.0cm has a mass of 0.3624g?

Homework Equations

D=m/V

The Attempt at a Solution

The problem I'm having is I don't understand what the problem is asking. How do you calculate the density of lithium metal with the diameter/length of a cylindrical wire and a the mass of the wire.

 

[hellomister]

You're given the dimensions of the rod, you can calculate the volume.

 

[Blueshift5]

The density of a substance is defined as its mass per unit volume. In this case, the density of lithium metal can be calculated using the formula D=m/V, where D is the density, m is the mass, and V is the volume.

In the given problem, we are given the mass of the cylindrical wire (0.3624g) and its dimensions (diameter = 2.40mm and length = 15.0cm). To calculate the volume of the wire, we can use the formula for the volume of a cylinder, V=πr^2h, where r is the radius (half of the diameter) and h is the height (length).

So, for the given wire, the radius would be 1.20mm (2.40mm/2) and the height would be 15.0cm. Plugging these values into the formula, we get V=π(1.20mm)^2(15.0cm) = 67.020mm^3.

Next, we can plug the values for mass (0.3624g) and volume (67.020mm^3) into the formula for density, D=m/V. This gives us a density of 0.0054g/mm^3.

However, the given units for density are g/cm^3, so we need to convert mm^3 to cm^3. 1mm^3 is equal to 0.001cm^3, so 67.020mm^3 would be equal to 0.06702cm^3.

Therefore, the density of lithium metal in g/cm^3 would be 0.0054g/0.06702cm^3 = 0.0807g/cm^3.