According to the given data;

`M =` mass of solid with water `= 103g`

`rho =` density of solid with water `= 2.65g/(cm^3)`

`w =` water content `= 15.6% = 0.156`

`V =` Volume of solid,water and air `= 56.6`

`M_D =` Dry mass of solid

`rho_D =` Dry density of solid

`M_D = M/(1+w)`

`M_D = 103/(1+0.156) = 89.1g`

`rho_D = rho/(1+w)`

`rho_D = 2.65/(1+0.156) = 2.292g/(cm^3)`

Volume of pure solid `= M_D/rho_D = 89.1/2.292 = 38.87cm^3`

Since the mass of air is negligible;

Mass of water = Total mass-mass of dry solid

Mass of water `= 103-89.1 = 13.9cm^3`

We know that density of water is `1g/cm^3`

Volume of water `= 1xx13.9cm^3 = 13.9cm^3`

Total volume = volume of solid+volume of water+volume of air

`56.6 = 38.87+13.9+V_A`

`V_A = 3.83cm^3`

**So the answers for the completed phase model diagram are;**

`M_D = 89.1g`

`M_W = 13.9g`

`M_A = 0g`

`V_D = 38.87cm^3`

`V_W =13.9cm^3`

`V_A = 3.83cm^3`

Water = 15.6% by weight

density of water =1

M= 103.0

Mass of water= 15.6% of 103 =16.07

Volume of water = 16.07

Volume of Solid= 56.6-16.07

=40.53

Mass of solid = 103-16.07

=86.93

`rho=86.93/40.53`

`=2.145`

After knowing M and denisty ,it is easy to calculate relative density by given formula.