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?

A) 2.5 g

B) 4.9 g

C) 9.8 g

D) 14 g

E) 28 g

Answer:

initially we have 500 mL of 0.118 M NaNO3
initial mol of NO3- = M*V
= 0.118 M * 0.500 L
= 0.0590 mol

final mol of NO3- = M*V
= 0.25 M * 0.500 L
= 0.125 mol

so,
mol of NO3- added = 0.125 mol - 0.0590 mol
= 0.066 mol

1 mol of Mg(NO3)2 has 2 mol of NO3- ion
So,
mol of Mg(NO3)2 added = mol of NO3- added / 2
= 0.066 mol / 2
= 0.033 mol

molar mass of Mg(NO3)2 = 148.3 g/mol

so,
mass of Mg(NO3)2 added = mol of Mg(NO3)2 added * molar mass
= 0.033 mol * 148.3 g/mol
= 4.89 g

Answer: 4.9 g