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13 January, 13:15

How many moles of solute particles are present in 100.0 ml of 2.50 m (nh4) 3po4?

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  1. 13 January, 14:47
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    The moles of solute particles that are present in 100.0 ml of 2.50M (NH4) 3Po4 is 0.25 moles

    calculation

    moles=volume in liters x molarity

    volume in liters=100/1000 = 0.1L

    molarity=2.50M

    moles is therefore = 2.50 x0.1=0.25moles
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