We know, Molarity= Moles/Volume(in L) Rearranging it we get, Number of Moles= Mo

Question

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

Explanation / Answer

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:

×

10

______mol of magnesium

_______ magnesium ions

_______mol of chloride

_________ chloride ions

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