Given 0.2 M solution of NaCl in water with a specific gravity of 1.35.
0.2 mol – 1000 mL
x mol – 800 mL
M(NaCl) = 58.44 g/mol
9.35 grams of NaCl are present in 800 mL of solution.
The formula for molality is
m = moles of solute / kilograms of solvent
0.148 mol/kg is the molality of the solution.