Question

The specific gravity of iron is $7.87$ and the density of water at ${4.00^0}C$ is $1.00gm{\left( {cm} \right)^{ - 3}}$ . You can use this information to find the density of iron. Find the volume occupied by $9.50g$ of iron?

Answer 1

The volume occupied by $9.50g$ of iron can be determined from the density of iron by dividing the given mass with the density of iron. The density of iron can be determined by the multiplication of specific gravity of iron and density of water at ${4.00^0}C$ .
${\rho _{Fe}} = SG \times {\rho _{{H_2}O}} \, at \, {4^0}C$
${\rho _{Fe}}$ is density of iron having to be calculated
$SG$ is specific gravity
${\rho _{{H_2}O}} \, at \, {4^0}C$ is the density of water at four degrees Celsius.

Complete answer:
Given that the specific gravity (SG) of iron is $7.87$
The density of water at ${4.00^0}C$ is $1.00gm{\left( {cm} \right)^{ - 3}}$
The density of iron is denoted by ${\rho _{Fe}}$ and has the formula of
${\rho _{Fe}} = SG \times {\rho _{{H_2}O}} \, at \, {4^0}C$
By substituting the values in the above formula, the value of density of iron will be ${\rho _{Fe}} = 7.87 \times 1 = 7.87g{\left( {cm} \right)^{ - 3}}$
The density gives the information that mass is exactly one unit of volume of that substance.
From the above value of the density of iron, one gram per centimetre cube occupies $7.87g$ .
$9.50g$ of iron occupies a volume of
$9.50 \times \dfrac{{1c{m^3}}}{{7.87g}} = 1.21c{m^3}$
Thus, the volume occupied by $9.50g$ of iron is $1.21c{m^3}$ when the specific gravity of iron is $7.87$ and the density of water at ${4.00^0}C$ is $1.00gm{\left( {cm} \right)^{ - 3}}$ .

Note:
The density is defined as the mass per unit volume and specific gravity of any substance can be calculated from the density and density of water at four degrees Celsius. As specific gravity is the ratio of density and reference material density. Generally used reference material is water at four degrees Celsius.