Question

Be sure to answer all parts.

How many moles and numbers of ions of each type are present in the following aqueous solution? 83.1 mL of 1.45 M magnesium chloride:

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10

______mol of magnesium

_______ magnesium ions

_______mol of chloride

_________ chloride ions

Answer 1

We know, Molarity= Moles/Volume(in L)

Rearranging it we get, Number of Moles= Molarity * Volume(in L)

Number of ions=moles*Avogadro's Constant, where Avogadro's Constant = 6.022*10²³

Using this information we do the calculations:

Mol of MgCl₂=1.45(mol/L)*83.1mL*0.001L/1mL=0.120mol

Now 1mol of MgCl₂ contains 1mol of Mg and 2 mol of Cl.(Looking the formula of Magnesium chloride for this).

So in 0.120 mol of MgCl₂ :

mol of Magnesium= 0.120 mol

mol of Chloride= 2*0.120=0.240mol

ions of Mg=0.120*6.022*10²³=7.23*10²²

ions of Cl=0.240*6.022*10²³=1.44*10²³