How many mL of a 0.131 M aqueous solution of iron(III) fluoride, , must be taken to obtain 5.24 grams of the salt?

Answer 1

How many moles is equal to 5.24 g of iron(III) fluoride (FeCl₃)? 5.24 g x 1 mole/162.2 g = 0.0323 moles

0.131 M = 0.131 moles/liter

(x liters)(0.131 mole/liter) = 0.0323 moles

x = 0.247 liters

xls needed = 0.247 liters x 1000 ml/liter = 247 mls (3 sig. figs.)