If the mole fraction of na2s in an aqueous solution is 0.125, what is the weight/weight % (percent by mass ) of na2s

Bthats what im doing to and i figured it out it is b

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Answerthe half-reaction having the greatest tendency to occur at the anode:     i₂(aq) + 2e⁻  ⇄ 2i⁻  ;   e = +0.53 v======oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be    i₂(aq) + 2e⁻  ⇄ 2i⁻    e = +0.53 vbecause its reduction potential is the lowest out of the three.

Moles = mass/molar mass the molar mass of c20 = 12.011*20 = 240.22 g per mole therefore, the number of moles is 20 g / 240.22 g per mol = 0.0833 moles.