Answer to Question #71141 in General Chemistry for Elana

Question

You need to make an aqueous solution of 0.229 M manganese(II) nitrate for an experiment in lab, using a 500 mL volumetric flask. How much solid manganese(II) nitrate should you add? Also, How many milliliters of an aqueous solution of 0.144 M lead acetate is needed to obtain 17.5 grams of the salt?

Answer 1

According to the definition, molar concentration of a substance in a solution is the ratio of number of the moles to the volume of the solution:

c=n/V.

The number of the moles is related to the mass with the molar mass:

n=m/M;m=n·M.
Thus, given the volume of the solution of manganese (II) nitrate, its concentration and molar mass (M(Mg(NO_3 )_2 ) =148.31 g mol^(-1)), we can calculate the mass of manganese (II) nitrate needed for the preparation :

m(Mg(NO_3 )_2=cVM=0.229 (mol L^(-1) )·0.5 (L)·148.31(g mol^(-1) )=16.9815 g.

The next question is the inverse problem. Knowing the concentration, mass and molar mass of lead acetate (325.29 g/mol), we can calculate the volume of the solution needed :

V=m/cM=(17.5 (g))/(0.144 (mol L^(-1) )·325.29 (g mol^(-1)))=0.3734 (L)=373.4 mL.

Answer: 16.9815 g of manganese (II) nitrate; 373.4 mL of the solution of lead acetate.