Answer to Question #157451 in Chemistry for zee

Question

Starting with the solid and adding water, how would you prepare 2.00 L of 0.685 M Ni(NO₃)₂?

Answer 1

The molecular mass of Ni(NO₃)₂ equals 182.7 g/mol.

According to the equation:

M = n / V = m / (Mr × V)

wher M - molar concentration (0.685 M), n - number of moles, Mr - molecular weight (182.7 g/mol), V - volume (2.00 L).

From here:

m = M × Mr × V = 0.685 M × 182.7 g/mol × 2.00 L = 250.3 g

As a result, to prepare the required solution, you need to take 250.3 g of Ni(NO₃)₂ and adjust it with water up to 2 L.