How many grams of Mg (NO3) 2 are in 160 ml of a 1.2M solution?

Given:
V solution (Mg (NO3) 2) = 160 ml = 0.16 l
Cm (Mg (NO3) 2) = 1.2 M

To find:
m (Mg (NO3) 2) -?

Decision:
1) Calculate the amount of Mg (NO3) 2 substance:
n (Mg (NO3) 2) = Cm (Mg (NO3) 2) * V solution (Mg (NO3) 2) = 1.2 * 0.16 = 0.192 mol;
2) Calculate the molar mass of Mg (NO3) 2:
M (Mg (NO3) 2) = Mr (Mg (NO3) 2) = Ar (Mg) * N (Mg) + Ar (N) * N (N) + Ar (O) * N (O) = 24 * 1 + 14 * 2 + 16 * 6 = 148 g / mol;
3) Calculate the mass of Mg (NO3) 2:
m (Mg (NO3) 2) = n (Mg (NO3) 2) * M (Mg (NO3) 2) = 0.192 * 148 = 28.42 g.

Answer: The mass of Mg (NO3) 2 is 28.42 g.