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

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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