Question

what is the concentration of K+ in 0.15M of K2S

Answer 1

K2S(aq) --> 2K+(aq) + S2-(aq)
1 mole of K2S produces 2 moles of K+,
concentration of K+ will be twice of K2S =
which is 0.30 M

Answer 2

since on dissolution of K2S
two K+ ions will be releases hence the conc is twice the conc of K2S

ie: 0.30 M

Answer 3

1.
K2S ---> 2K+ + S2-
So,
concentartion of K+ = 2* concentration of K2S
                                        = 2*0.15
                                         =0.3 M

2.
[Na3PO4] = [Na+]/3
                     =1.5/3
                     =0.5 M
V= 525 mL = 0.525 L
Number of moles of Na3PO4 = M*V = 0.5*0.525 = 0.2625 mol
Molar mass of Na3PO4 = 164 g/mol
mass of Na3PO4 = number of moles * molar mass
                                  = 0.2625 * 164
                                   = 43.05 g
Answer: 43.05

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