Question

When 0.064L of a solution containing 1.93M BaF₂ is diluted to 0.2L, what is the concentration of BaF₂ in the new solution?

Answer 1

We need to apply dilution law, which is based on the mass conservation principle

initial mass = final mass

this apply for moles as weel ( if there is no reaction, which is the case )

mol of A initially = mol of A finally

or, for this case

moles of A in stock = moles of A in diluted solution

Recall that

mol of A = Molarity of A * Volume of A

then

moles of A in stock = moles of A in diluted solution

Molarity of A in stock * Volume of A in stock = Molarity of A in diluted solution* Volume of A in diluted solution

Now, substitute known data

M1V1 = M2V2

0.064*1.93 = 0.2*M2

M2 = 0.064*1.93 /0.2

M = 0.6176 mol of BaF2/L