18. The chloride in a 0.12-g sample of 95% pure MgCl2 is to be precipitated...
Solutions, Solubility and Concentration. See attached file for full problem description.
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18. Write the reaction
MgCl2 + AgNO3 = 2AgCl + Mg(NO3)2
MgCl2 = 24 + 71 = 95 (g/mol)
AgNO3 = 108 + 14 + 48 = 170 (g/mol)
Number of mole of MgCl2 = 0.12*0.95/95 = 0.0012 (mol)
So the number of mole of AgNO3 required to react is also 0.0012 mol.
Volume of ...
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