Question

35. How many moles of solute particles are present in 100.0 mL of 2.50 M (NH4)3PO4?

A) 0.100 mol

B) 0.250 mol

C) 0.500 mol

D) 0.750 mol

E) l.00 mol

Ans: E.. how do you get the answer?

Answer 1

yes option E correct:

Molarity = no of moles of solute x1000/ V(ml)

no of moles of solute= M x V /1000

                                   = 2.5 x 100/1000

                                    = 0.25

there are 4 solute particles (3+1) present those are 3 NH4⁺   and 1 PO4^-3 .  

moles of solute particles = 4 x 0.25 = 1.00mol