Question

What mass of Ba(NO3)2 would you need to make a 500 mL solution with [NO3-]=0.125 M?

The answer is 8.17 g

Please explain each equation and why with a step by step guide on how to arrive at 8.17 g. Thank you!

Answer 1

1 mol of Ba(NO3)2 has 2 moles of NO3-

So,

[Ba(NO3)2] = [NO3-]/2

= (0.125)/2

= 0.0625 M

volume , V = 500 mL

= 0.5 L

we have below equation to be used:

number of mol,

n = Molarity * Volume

= 0.0625*0.5

= 3.125*10^-2 mol

Molar mass of Ba(NO3)2 = 1*MM(Ba) + 2*MM(N) + 6*MM(O)

= 1*137.3 + 2*14.01 + 6*16.0

= 261.32 g/mol

we have below equation to be used:

mass of Ba(NO3)2,

m = number of mol * molar mass

= 3.125*10^-2 mol * 261.32 g/mol

= 8.17 g

Answer: 8.17 g