Question

What mass (in grams) of Mg(NO3)2 is present in 183 mL of a 0.400 M solution of Mg(NO3)2 ?

0.0732 g
10.9 g
6.32 g
67.8 g

Answer 1

B.) 10.9 g

Step-by-step explanation:

To find the mass of Mg(NO₃)₂, you need to (1) convert the volume from mL to L, then (2) calculate the amount of moles (using the molarity equation), and then (3) convert moles to grams (using the molar mass). It is important to arrange the conversions/ratios in a way that allows for the cancellation of units.

(Step 1)

1,000 mL = 1 L

183 mL 1 L
-------------- x ----------------- = 0.183 L
1,000 mL

(Step 2)

Molarity = moles / volume (L) <----- Molarity equation

0.400 M = moles / 0.183 L <----- Insert variables

0.0732 = moles <----- Multiply both sides by 0.183

(Step 3)

Atomic Mass (Mg): 24.305 g/mol

Atomic Mass (N): 14.007 g/mol

Atomic Mass (O): 15.999 g/mol

Molar Mass (Mg(NO₃)₂): 24.305 g/mol + 2(14.007 g/mol) + 6(15.999 g/mol)

Molar Mass (Mg(NO₃)₂): 148.313 g/mol

0.0732 moles Mg(NO₃)₂ 148.313 g
--------------------------------------- x --------------------- = 10.9 g Mg(NO₃)₂
1 mole