An aqueous solution is made by dissolving 29.4 grams of nickel(II) nitrate in 403 grams of water.
The molality of nickel(II) nitrate in the solution is m
Solution:
nickel(II) nitrate = Ni(NO3)2
The molar mass of Ni(NO3)2 is 182.703 g/mol
Therefore,
Moles of Ni(NO3)2 = [29.4 g Ni(NO3)2] × [1 mol Ni(NO3)2 / 182.703 g Ni(NO3)2] = 0.1609 mol Ni(NO3)2
Kilograms of water = (403 g) × (1 kg / 1000 g) = 0.403 kg
Molality = Moles of solute / Kilograms of solvent
Therefore,
Molality of Ni(NO3)2 solution = (0.1609 mol) / (0.403 kg) = 0.399 mol/kg = 0.4 m
Molality of Ni(NO3)2 solution = 0.4 m
Answer: The molality of nickel(II) nitrate in the solution is 0.4 m
Comments
Leave a comment