Question

determine the concentration (in mol dm^-3) of the resulting solution when 350cm^3 of 0.800 mol dm^-3 hydrochloric acid solution is mixed with 200cm^3 of 1.30 mol dm^-3 hydrochloric acid solution

Answer 1

This per means Volume of Solution A = 350 cm = 0.35 dm Conc. ne on of Solntion A= 0.800 mol/dm means that there are 0.800 moles dm of solution and 6. Namber of moles in 0.35 dim of alb solution A = 0.35 x 0.800 mol/dm² dm3 1 0 :28 mol Volume of Solution B= 200cm3 = 0.2 dm * Conc. 1 4 = 1:30 mol/dm3 Number of moles in oglan of 8trat Solution A = 0.2 dmx 1.30 mol/dm3 20.26 mot. After mining of solution A and B Total volume - (0.35 +012) and = oiss dm Total number of molers in 0.55 dm of the mixed solution - 10.28 + 0.26) mot. = 0.54 mol. 0.982 mol/dme :: Number of moles per din of solution = 0.54 mot = oiss din Hence the concentration of the solution = 0.982 mol/dm3 resulting