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)

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