Chemistry related question?
Also,
You prepare a solution by placing 42.5 g of CaCl2 in 500 mL of solution. Then, dilute that solution to 0.25M. How much of the 0.25M solution could be made?
Answer:
You need to know the density of the BaBr2 solution to get the answer.
750 mL x density (g/ml) x .175 = grams of BaBr2
42.5 gm of CaCl2 is 42.5 / 110.99 = 0.3829 moles
(110.99 is the MW of CaCl2)
Liters x Molarity = Moles
Liters x 0.250 = 0.3829
Liters = 1.532 = 1,532 mL
Well if you have a 17.5% solution the you have 17.5g/100mL or 175g/1000mL So if you have 750mL of solution then you have:
175g*0.75=131.75grams
42.5g*(1moleCaCl2/110.984g)=0.
Then you use the equation Mconc*Vconc=Mdiluted*Vdiluted
(0.766M*0.5L)=(0.25M*x) Solve for x
x=1.532Liters so approximately 1.5Liters
The answers post by the user, for information only, FunQA.com does not guarantee the right.
More Questions and Answers: