# Which of acid aqueous solution 1) to 4) is non-volatile one?

I am working on my scholarship exam practice assuming high school chemistry knowledge and struggling at how I can rationalize this question.

Which of acid aqueous solution 1) to 4) is non-volatile one?
1) $$\ce{CH3COOH}$$
2) $$\ce{H2SO4}$$
3) $$\ce{HCl}$$
4) $$\ce{HF}$$

The answer key provided is 2) $$\ce{H2SO4}$$ but I wonder, in the exam condition, how should I solve this rationally? What thought should I have on this to determine the volatility of the compound? Please help.li

• Chemistry is not quite like math. You can't expect to deduce everything rationally. That said, I'd think of the bonds holding these molecules together (what are they, BTW?), and compare the compounds along these lines. – Ivan Neretin Jun 20 '19 at 9:45
• I found this question very broad since the exam does not specify the scope of content and it is from other country (Japan) which has different curriculum so I tried to find some rules of thumb to tackle such problems. Anyway, I think yes, $H_2SO_4$ has two hydrogen bonds with two additional polar $O$'s while others have less. For example, $CH_3COOH$ and $HF$ only contain a hydrogen bond while $HCl$ not. Perhaps molecular weight should be taken into account too. – Trey Anupong Jun 20 '19 at 10:00
• Forget molecular weight; hydrogen bonds are quite enough of a reason. See, despite what I said in the first comment, it is possible to solve this one logically after all, and you just did that yourself. – Ivan Neretin Jun 20 '19 at 10:11
• @TreyAnupong Please have a look on chemistry.meta.stackexchange.com/questions/86/…, with mhchem-syntax (equally available for LaTeX) input and rendering of chemical equation and formulae is greatly improved and may be equally used in questions, answers and commentaries. About content of your question, at grammar school-level we had little booklets with math / phys / chem relevant tables repeating textbook content, including some about typical chemicals and their properties (molar mass, melting / boiling point). Didn't you too? – Buttonwood Jun 20 '19 at 20:17