how many "particles in solution" will form from each sucrose (C12H22O11) molecule added to water? Explain.

Question

I don't understand where I am supposed to start... What do I need to know to do this problem?
thank you

Answer 1

You need to know that C12H22O11 is sucrose. It is held together by covalent bonds and doesn't ionize when placed in water. Therefore, sucrose will dissolve in water forming one particle per molecule. Something like NaCl, an ionic compound, dissolves in water forming two particles per molecule; i.e., a Na^+ and a Cl^-. MgCl2, an ionic compound, dissolves in water forming three particles per molecule MgCl2; i.e., one Mg^2+ ion and two Cl^- ions. Covalent molecules other than sugar, if they dissolve in water, dissolve as the molecule; therefore there is only one particle per molecule. I hope this helps.

Answer 2

Okay thank you for the explanation. For soduim sulfide(NaS), will it form 3 particles in solution? Na+1 and S-2 so would that give you 3?

For future reference, for these problems should i first see what type of bond it creates, then if it dissolves in water to get my answer?

Answer 3

Sodium sulfide does, indeed, form three particles, but your explanation is not right. Sodium sulfide is Na2S (you wrote NaS). When it dissolves it forms 2 Na^+ and 1 S^2- for three particles. (Don't confuse the number of ions with the charge. The sulfide ion is -2 charge but it's only one ion.)
Na2S ==> 2Na^+ + S^2-

For the futures question, I think that will work.