What  volume  of  a  0.240  m  solution  of  barium  nitrate  is  needed  to  prepare  0.500  l  a  solution  that  is  0.0800  m  in  nitrate  ion?

Formula for Barium Nitrate = Ba(NO3)2

Thus based on stoichiometry:

1 mole of Ba(NO3)2 contains 2 moles of NO3-

Therefore, concentration of nitrate ion NO3- would be = 2*0.240 = 0.480 M

Use the relation:

V1M1 = V2M2

V1 = V2M2/M1 = 0.500 L * 0.0800/0.480 = 0.0833 L

Thus, 0.0833 L or 83.3 ml solution of Ba(NO3)2 would be required.

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