Answer
The specific gravity of the mixture is 0.91
Work Step by Step
We can find the mass $M_a$ of the antifreeze solution.
$M_a = (0.80)(1.000\times 10^3~kg/m^3)(10^{-3}~m^3/L)(4.0~L)$
$M_a = 3.2~kg$
We can find the mass $M_w$ of the water.
$M_w = (1.000\times 10^3~kg/m^3)(10^{-3}~m^3/L)(5.0~L)$
$M_a = 5.0~kg$
The total mass of the mixture is 8.2 kg. We can find the density of the mixture.
$\rho = (\frac{8.2~kg}{9.0~L})~(\frac{1~L}{10^{-3}~m^3})$
$\rho = 0.91\times 10^3~kg/m^3$
The specific gravity of the mixture is 0.91