Answer:
1) 0.9972 g/cm³.
2) 77.62 g.
3) 25.01 cm³.
4) 7.11 cm³
5) 17.9 cm³.
6) 4.34 g/cm³
Explanation:
Hello.
In this case, for this gravimetry problem, we proceed as follows:
1) Since the temperature is 23.75 °C the density computed via lineal interpolation is:
[tex]\rho=1.004g/cm^3+23.75\°C*-0.000273\frac{g/cm^3}{\°C} =0.9972g/cm^3[/tex]
2) The mass of the solid is computed via the mass of the empty flask and the mass of the flask with the solid:
[tex]m_{solid}=135.95g-58.33g\\\\m_{solid}=77.62g[/tex]
3) In this case, since the volume of the flask equals the volume of the flask and the water because the volume of liquid water equals the volume of the container at which it is, we compute the volume of the flask considering both the mass and density of water:
[tex]m_{water}=83.27g-58.33g=24.94g\\\\V_{water}=\frac{24.94g}{0.9972g/cm^3}\\ \\V_{water}=25.01cm^3[/tex]
Thus, the volume of the flask is 25.0 cm³.
4) Here, we compute the volume of water when the water and the solid are inside the flask:
[tex]m_{water}=143.04g-135.95g=7.09g\\\\V_{water}=\frac{7.09g}{0.9972g/cm^3} =7.11cm^3[/tex]
Thus, the volume of the flask not occupied by the solid which corresponds to the water is 7.11 cm³.
5) Considering the previous volume of the flask, the volume of the solid is:
[tex]V_{solid}=25.01 cm^3-7.11cm^3\\\\V_{solid}=17.9cm^3[/tex]
6) Finally, the density of the solid is:
[tex]\rho _{solid}=\frac{77.62g}{17.9cm^3}\\ \\\rho _{solid}=4.34g/cm^3[/tex]
Regards.
