Chem question please help

Question

what mass of magnesium bromide would be required to prepare 720.ml of a 0.0939 M aqueous solution?

Answer 1

molar mass magnesium bromide (MgBr₂) = 184 g/mole

0.0939 mol/L x 0.720 L = 0.06761 moles MgBr₂ required

0.06761 moles x 184 g/mole = 12.4 grams MgBr₂ required (to 3 significant figures)

Answer 2

Molarity, M = mole/L

convert 720 mL to L (1000mL are in 1 L)

 

(0.0939 mol/L)×(0.72 L) = 0.0676 mol MgBr2

 

**Note that the given quantities in the question have 3 significant figures so round answers to 3 sigfigs)

 

Now that we have the mole magnesium bromide required. Use the Molecular mass of magnesium bromide to calculate mass required.

 

Molecular mass of magnesium bromide (MgBr2) = (24.3 g/mol Mg) + 2(79.9 g/mol Br) = 184 g/mol MgBr2

 

(0.0676 mol MgBr2) × (184 g/mol MgBr2) = 12.4 g MgBr2