Given that the density of heptane is
[tex]d_h=\frac{0.684g}{mL}[/tex]The mass of heptane is
[tex]m_h=31\text{ g}[/tex]The density of water is
[tex]d_w=\frac{1g}{mL}[/tex]The mass of water is
[tex]m_w=37\text{ g}[/tex]The volume of heptane will be
[tex]\begin{gathered} V_h=\frac{m_h}{d_h} \\ =\frac{31}{0.684} \\ =45.32\text{ mL} \end{gathered}[/tex]The volume of water will be
[tex]\begin{gathered} V_w=\frac{m_w}{d_w} \\ =\frac{37}{1} \\ =37\text{ mL} \end{gathered}[/tex]Thus, the volume of heptane is 45.32 mL and the volume of water is 37 mL.
The total volume of liquid in the cylinder will be
[tex]\begin{gathered} V=V_h+V_w \\ =45.32+37 \\ =82.32\text{ mL} \end{gathered}[/tex]The total volume of liquid in the cylinder will be 82.32 mL.