Calculate the molality of 1 L solution of 80% H₂SO₄. (w/V), given that the density of the solution is 1.80 g mL^-1.

${\mathrm{W}}_{{\mathrm{H}}_2{\mathrm{SO}}_4}$ =800 c in 10oo mL
volume of solution = 1000 mL
Weight of solution = 1000.0 x 1.80
= 1800 g
weight of water =1800 - 800 =1000 g
$\mathrm{Molality} =\frac{800\times 1000}{98\times 1000}=8.16\mathrm{mol} {\mathrm{kg}}^{-1}$