Determine the molarity of Ni2+ ions.

Question

Determine the molarity of Ni²⁺ ions in solution when 5.57×10^-1 g of nickel(II) nitrate is added to enough water to prepare 3.58×10² mL of the solution.

Answer 1

Solution

mass (g) = 5.57 x 10^-1g = 0.557 g

molar mass (g/mol) Ni(NO₃)₂ = 183 g/mol

volume of solution = 3.58 x 10² mL = 358 mL = 0.358 L

mol = mass (g) / molar mass (g/mol) = 0.557 g /183 g/mol = 0.00304 mol

Molarity (M) = moles (n) / volume (L)

Molarity (M) = 0.00304 mol / 0.358 L = 0.00858 mol/L = 0.00858 M

Answer 2

Molar mass Ni(NO₃)₂ = 183g/mol

moles Ni(NO₃)₂ = 0.557 g x1 mol/183 g = 0.00304 moles

Since 1 moles Ni(NO₃)₂ yields 1 mole Ni²⁺ so we have 0.00304 moles Ni²⁺

We have a volume of 358 ml or 0.358 L of solution. Since molarity is defined as moors /liter of solution…

molarity of Nii²⁺= 0.00304 mols/0.358 L = 0.00858 M