How many grams of $CuF_2$ are needed to make a 600 mL solution with a molarity of 0.250 M?

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Answer 1 · 2016-08-18T07:57:39

Approx. $15*g$

$"Concentration"$ $=$ $"number of moles"/"volume of solution"$

Thus $0.250*mol*L^-1$ $=$ $n_"copper fluoride"/(0.600*L)$

And $n_"copper fluoride"=0.250*mol*cancel(L^-1)xx0.600*cancelL$

$=$ $0.150*mol" copper fluoride"$

$"Mass of cupric fluoride"$ $=$ $0.150*molxx101.54*g*mol^-1$ $=$ $??g$

How many grams of CuF_2CuF_2 are needed to make a 600 mL solution with a molarity of 0.250 M?