Question

Iron has a density of 7.86 g/cm3. Calculate the volume (in dL) of a piece of iron having a mass of 4.07 kg . Note that the density is provided in different units of volume and mass than the desired units of volume (dL) and the given units of mass (kg). You will need to express the density in kg/dL (1 cm3 = 1 mL) before calculating the volume for the piece of iron.

Answer 1

Answer:

The volume of the piece of iron is 5.18dL.

Explanation:

The density (ρ) is equal to the mass (m) divided the volume (V).

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

If we rearrange it, we have:

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

To express the volume in dL we will need the following relations:

• 1 dL = 0.1 L
• 1 kg = 10³ g
• 1 cm³ = 1 mL
• 1mL = 10⁻³L

Then,

$\rho = \frac{7.86g}{cm^{3} } .\frac{1cm^{3} }{1mL} .\frac{1mL}{10^{-3}L } .\frac{1kg}{10^{3}g } =7.86kg/L$

Finally,

$V=\frac{m}{\rho }=\frac{4.07kg}{7.86 kg/L} =0.518L.\frac{1dL}{0.1L} =5.18dL$

Answer 2

Answer:

5.178 dL.

Explanation:

Density of iron = 7.86g/cm^3.

1 dL = 1/10 of a liter.

1 dL = 100 mls

and 1 gm = 1/1000 of a kg.

So 7.86 g/cm^3 =  7.86 / 1000 =  0.00786 kg / cm^3

= 0.00786 * 100

= 0.786 kg / dL  is the density.

Volume = mass / density

= 4.07 / 0.786

= 5.178 dL.