Question

What mass (in grams) of MgBr2 would be present in 23.7 mL of 0.875 M MgBr2 solution? Give your answer to 2 decimal places. (You’ll need your periodic table)

Answer 1

A) 0.875 M of MgBr2 is 0.875 mol/L

B) 23.7 mL = 0.0237 L


0.875 mol => 1 L
x mol => 0.0237 L

Cross multiply
1x = 0.875 × 0.0237
x = 0.0207 mol

>> In 23.7 mL of 0.875 M MgBr2 solution there is 0.0207 moles of MgBr2




C) Molar mass of MgBr2 (*) = 24.305 + (2 × 79.904) = 184.113 g/mol


184.113 g => 1 mol
x g => 0.0207 mol

Cross multiply
1x = 184.113 × 0.0207
x = 3.8111 g

>> 0.0207 moles of MgBr2 is equivalent to 3.81 g of MgBr2
>> In 23.7 mL of 0.875 M MgBr2 solution there is 3.81 g of MgBr2



(*) Use your periodic table