Given a $1 \cdot L$ $1L$ volume of solution that is $1.25 \cdot m o l \cdot L^{- 1}$ $1.25mol*L^-1$ with respect to $M g C l_{2}$ $MgCl_2$ , there are...? $a. 2 moles of ions in solution.$ $"a. 2 moles of ions in solution."$ $b. 1.25 moles of ions in solution.$ $"b. 1.25 moles of ions in solution."$ $c. 2.5 moles of ions in solution.$ $"c. 2.5 moles of ions in solution."$ $d. 2 moles of ions in solution.$ $"d. 2 moles of ions in solution."$

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Given a $1 \cdot L$ $1L$ volume of solution that is $1.25 \cdot m o l \cdot L^{- 1}$ $1.25mol*L^-1$ with respect to $M g C l_{2}$ $MgCl_2$ , there are...? $a. 2 moles of ions in solution.$ $"a. 2 moles of ions in solution."$ $b. 1.25 moles of ions in solution.$ $"b. 1.25 moles of ions in solution."$ $c. 2.5 moles of ions in solution.$ $"c. 2.5 moles of ions in solution."$ $d. 2 moles of ions in solution.$ $"d. 2 moles of ions in solution."$

Given a $1 \cdot L$ $1L$ volume of solution that is $1.25 \cdot m o l \cdot L^{- 1}$ $1.25mol*L^-1$ with respect to $M g C l_{2}$ $MgCl_2$ , there are...
$a. 2 moles of ions in solution.$ $"a. 2 moles of ions in solution."$
$b. 1.25 moles of ions in solution.$ $"b. 1.25 moles of ions in solution."$
$c. 2.5 moles of ions in solution.$ $"c. 2.5 moles of ions in solution."$
$d. 2 moles of ions in solution.$ $"d. 2 moles of ions in solution."$

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Answer 1 · 2017-05-03T05:17:01

$Option c............$ $"Option c............"$

Explanation:

The $mole$ $"mole"$ , $N_{A}$ $N_A$ , is simply a number. Admittedly, it is a very large number, $N_{A} = 6.022 \times 10^{23} \cdot m o l^{- 1}$ $N_A=6.022xx10^23*mol^-1$ .

And if you have a mole of stuff, you have $6.022 \times 10^{23}$ $6.022xx10^23$ INDIVIDUAL ITEMS of that stuff. Here you have a concentration of $1.25 \cdot m o l \cdot L^{- 1}$ $1.25*mol*L^-1$ $M g C l_{2}$ $MgCl_2$ . In $1 \cdot L$ $1*L$ of solution there are $1.25 \cdot m o l$ $1.25*mol$ $M g^{2 +}$ $Mg^(2+)$ ions, BUT ALSO $2 \times 1.25 \cdot m o l$ $2xx1.25*mol$ $C l^{-}$ $Cl^-$ ions.....

And thus in total you have $2.5 \times N_{A}$ $2.5xxN_A$ $C l^{-}$ $Cl^-$ 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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