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21 December, 12:27

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?

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  1. 21 December, 12:45
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    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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