(II) If 4.0 L of antifreeze solution (specific gravity = 0.80) is added to 5.0 L of water to make a 9.0 L mixture, what is the specific gravity of the mixture?

The specific gravity of the mixture is described as the ratio of the density of the given mixture to water density. The antifreeze density depends on the specific gravity of the antifreeze and the density of the water.

The given data can be listed below as,

• The specific gravity of the antifreeze solution is S.G_af=0.80.
• The volume of the antifreeze solution is, V_af = 4L.
• The volume of the water is, V_w = 5L.
• The volume of the mixture is, V_m = 9L.

The specific gravity of the mixture can be expressed as,

\begin{aligned}S.{G_{\rm{m}}}=\frac{{{\rho_{\rm{m}}}}}{{{\rho_{\rm{w}}}}}\\=\frac{{{{\left({\frac{m}{V}}\right)}_{\rm{m}}}}}{{{\rho_{\rm{w}}}}}\\=\frac{{{m_{\rm{m}}}}}{{{V_{\rm{m}}}{\rho _{\rm{w}}}}}\end{aligned}

Here,ρ_m is the density of the mixture, ρ_w is the density of the water, m_m is the mass of the mixture, V_m is the volume of the mixture.

This can be written as,

\begin{aligned}S.{G_{\rm{m}}}=\frac{{{m_{{\rm{af}}}}+{m_{\rm{w}}}}}{{{\rho_{\rm{w}}}{V_{\rm{m}}}}}\\=\frac{{{\rho_{{\rm{af}}}}{V_{{\rm{af}}}}+{\rho_{\rm{w}}}{V_{\rm{w}}}}}{{{\rho_{\rm{w}}}{V_{\rm{m}}}}}\\=\frac{{S.{G_{{\rm{af}}}}\times{\rho _{\rm{w}}}{V_{{\rm{af}}}}+{\rho_{\rm{w}}}{V_{\rm{w}}}}}{{{\rho_{\rm{w}}}{V_{\rm{m}}}}}\\=\frac{{S.{G_{{\rm{af}}}}\times{V_{{\rm{af}}}}+{V_{\rm{w}}}}}{{{V_{\rm{m}}}}}\end{aligned}

Here, m_af is the mass of the antifreeze solution, V_af is the volume of the antifreeze solution, V_w is the volume of the water, m_w is the mass of the water.

Substitute the values in the above equation.

\begin{aligned}S.{G_{\rm{m}}}=\frac{{0.80\times4{\rm{L}}+5{\rm{L}}}}{{9{\rm{L}}}}\\=0.911\end{aligned}

Thus, the specific gravity of the mixture is 0.911.