Given a $1*L$ volume of solution that is $1.25*mol*L^-1$ with respect to $MgCl_2$, there are...? $"a. 2 moles of ions in solution."$ $"b. 1.25 moles of ions in solution."$ $"c. 2.5 moles of ions in solution."$ $"d. 2 moles of ions in solution."$

Md Ariful Islam

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Md Ariful Islam

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The $"mole"$, $N_A$, is simply a number. Admittedly, it is a very large number, $N_A=6.022xx10^23*mol^-1$.

And if you have a mole of stuff, you have $6.022xx10^23$ INDIVIDUAL ITEMS of that stuff. Here you have a concentration of $1.25*mol*L^-1$ $MgCl_2$. In $1*L$ of solution there are $1.25*mol$ $Mg^(2+)$ ions, BUT ALSO $2xx1.25*mol$ $Cl^-$ ions.....

And thus in total you have $2.5xxN_A$ $Cl^-$ per litre of this solution. This is a basic, underlying principle of chemistry, and if you can develop this idea of chemical equivalence, where MASS and CONCENTRATION corresponds to numbers of molecules and atoms, you will save yourself a lot of trouble,

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