Answer to Question #307649 in Chemistry for jack

Question

If a chemist has a stock solution of BF₃ that is 8.5 M and would like to make 325 mL of 2.0 M BF₃, how would she do it?

Answer 1

Solution:

C₁ = 8.5 M

V₁ = unknown

C₂ = 2.0 M

V₂ = 325 mL

Therefore,

V₁ = (C₂ × V₂) / C₁

V₁ = (2.0 M × 325 mL) / (8.5 M) = 76.47 mL

V₁ = 76.47 mL

Volume of stock BF₃ solution = 76.47 mL

Volume of solvent = 325 mL − 76.47 mL = 248.53 mL

Therefore, to make 325 mL of 2.0 M BF₃ solution, it is necessary to take 76.47 mL of 8.5 M a stock BF₃ solution and dissolve it in 248.53 ml of the solvent.