Answer to Question #295183 in General Chemistry for sss

Question

What volume of a 0.300 M solution of aluminum fluoride would be required to obtain 14.0 g of aluminum fluoride

Answer 1

Molar mass of aluminium fluoride $=83.9767g/mol$

$\frac{V}{1000}×0.3=\frac{14}{83.9767}$

$V=555.71cm^3$