How many grams of solid magnesium nitrate (148.3 g/mol) should be added to 500 mL of 0.118 M of sodium nitrate in order to produce an aqueous 0.25 M solution of nitrate ions?

 

Answer:

500 mL of 0.118 M of sodium nitrate = 0.500*0.118 =0.059 mole.

Mg(NO3)2 ....> Mg2+ + 2 NO3-

here volume chage negligible after mixing Mg(NO3)2.

so total volume = 500 ml

500 ml of 0.25 M solution of nitrate ions = 0.500*0.25 = 0.125 mole

so nitrate ion come from Mg(NO3)2 = (0.125-0.059) = 0.066 mole

so mole Mg(NO3)2 = 0.066/2= 0.033 mole = 0.033*148.3 = 4.89 gm (answer)