The density of hematite ore is 4.78 g/cm³.
Let the mass of the ore be m₁ , the volume of the flask V, total mass of water and ore M, mass of water be m₂, density of ore ρ₁ and the density of water be ρ₂.
Density of a substance is defined as the mass per unit volume.
The mass of water in the flask is given by,
[tex] M-m_1= (109.3 g)-(70.7 g)=38.3 g [/tex]
Calculate the volume V₂ of the water in the flask.
[tex] V_2=\frac{m_2}{\rho_2} =\frac{38.3 g}{0.997 g/cm^3} = 38.4cm^3 [/tex]
Calculate the volume V₁occupied by the ore in the flask.
[tex] V_1=V-V_2=(53.2 cm^3)-(38.4cm^3)=14.8cm^3 [/tex]
1 ml is equivalent to 1 cm³.
Calculate the density of the ore.
[tex] \rho_1=\frac{m_1}{V_1}=\frac{70.7 g}{14.8 cm^3} = 4.78g/cm^3 [/tex]
The density of the hematite ore is found to be 4.78 g/.cm³