What mass (in grams) of mg(no3)2 is present in 151 ml of a 0.350 m solution of mg(no3)2? a) 7.84 gb) 0.0529 gc) 4.56 gd) 64.0 g

Answer is: A) 7.84 g.

V(Mg(NO₃)₂) = 151 mL ÷ 1000 mL/L.

V(Mg(NO₃)₂) = 0.151 L; volume of the magnesium nitrate.

c(Mg(NO₃)₂) = 0.352 M; molarity of the solution.

n(Mg(NO₃)₂) = V(Mg(NO₃)₂) · c(Mg(NO₃)₂).

n(Mg(NO₃)₂) ) = 0.151 L · 0.352 mol/L.

n(Mg(NO₃)₂) = 0.0531 mol; amount of the substance.

M(Mg(NO₃)₂) = Ar(Mg) + 2Ar(N) + 6Ar(O) · g/mol.

M(Mg(NO₃)₂) = 24.3 + 2·14 + 6·16 · g/mol.

M(Mg(NO₃)₂) = 148.3 g/mol; molar mass.

m(Mg(NO₃)₂) = n(Mg(NO₃)₂) · M(Mg(NO₃)₂).

m(Mg(NO₃)₂) = 0.0531 mol · 148.3 g/mol.

m(Mg(NO₃)₂) = 7.84; mass of magnesium nitrate.

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