Question

how would you prepare a stock solution of 0.100 M solution of MgCl2 in 500 mL? Using the stock solution, how would you make a 0.078 M solution in 25.00 mL?

Answer 1

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of mg Cl O number of moles required Molarity & Volume of sole.. = 0.1008 0.5 = 0.05 moles .: mass of my cla required = moles of Mg Cbx molar mass = 0.05moles X 95 g/mol = 4.57g Ans Using 'V'volume of stock solution to prepare 0.078 M solution in @25 ml " number of moles of Mylla will be same - 0.100 x V = 0.078 x 25ml >V= 19.5 ml) Ang