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Magnesium phosphate is used to minimize the effects of vitamin E deficiency. A 3.80L solution has a magnesium ion concentration of 0.798mol/L. Calculate the mass of solute needed to prepare this solution.

1 month ago

Answered By Mandy H

Start with a dissociation equation for magnesium phosphate. Dissociation occurs when an ionic compound breaks up into its ions. 

Mg3(PO4)2 (s) --> 3Mg2+(aq) + 2PO43-(aq)

The next step is to list all the information you know from the question:

Volume = 3.80 L

Concentration = 0.798 mol/ L Mg2+

m = ? g Mg3(PO4)

Molar mass = 262.87 g/mol Mg3(PO4)2

There are two ways to solve this question. 

The first way is to use unit analysis (arrange the question so that all the units will 'cancel' out):

3.80 L x (0.798 mol Mg2+/L) x (1 mol Mg3(PO4)2 / 3 mol Mg2+) x (262.87 g/mol Mg3(PO4)2) = 265.708...g Mg3(PO4)2

The second way is to use formulas: 

1.) concentration = moles/ volume 

Manipulate this formula to find moles (moles = concentration x volume) 

moles of Mg2+ = 3.80 L x 0.798 mol Mg2+/L = 3.0324 mol Mg2+

2.) mole ratio (from the balanced chemical formula) 

3.0324 mol Mg2+ x (1 mol Mg3(PO4)2 / 3 mol Mg2+)  = 1.0108 mol Mg3(PO4)2

3.) number of moles = (mass of substance/ molar mass) 

Manipulate the formula to find mass of substance (mass of substance =  number of moles x molar mass)

mass of Mg3(PO4)2 =  1.0108 mol Mg3(PO4)x 262.87 g/mol Mg3(PO4)=265.708...g Mg3(PO4)2

The final answer is the mass of magnesium phosphate required is 266 g Mg3(PO4)(this is considering the fact there is three significant digits from the question so the question is rounded to three digits).