Doubt in the word infinite dilution [closed]

At times I find the concentrations being expressed as weight percentage. However I do not understand why molarity is not chosen.

Secondly, which close reference to the problem that I have been facing this is a follow up question:

The change of standard state from pure liquid to 1 wt % for Si dissolved in liquid Fe at 1873 K is expressed as: $$\ce{Si(liq)=Si(1 wt\%)}$$. Given that the activity coefficient of Si at infinite dilution is $$10^{-3}$$, what is the standard Gibbs free energy change (in kJ) for this equilibrium.

Doubt What is so special here about the word "infinite dilution?"

closed as too broad by Mithoron, Mathew Mahindaratne, A.K., Todd Minehardt, Nuclear ChemistSep 8 at 17:31

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• If a substance has unknown molecular weight, molarity is not feasible. But if the substance can be weighed, then the weight percent option is viable. – Ed V Sep 4 at 22:15
• Also w% either referred to solution or solvent is very practical WHEN the solution is very diluted and even more when the solvent is water. This is what is easily forgotten even by experienced scientists leading to almost useless protocols. 90% of the protocols I see in interdisciplinary research can be used only assuming an incorrect compilation of them. This is just a rant because something so easy should be not accepted. For the answer see both mothgod and M. Farooq. – Alchimista Sep 5 at 9:58

So then, the question is essentially asking what the Gibbs' Free Energy change is when dissolving liquid $$\ce{Si}$$ in liquid $$\ce{Fe}$$ to make a $$1\%$$ solution. Since you are working with a solution at infinite dilution, the problem allows you to ignore any additional interactions between dissolved Si particles, and can use the activity given in the problem.