Question

If volumes are additive and 280.0 mL of 2.40 M potassium bromide is mixed with 120 mL of a calciumbromide solution to give a new solution in which [Br-1] is 1.85 M, what is the concentration of the calciumbromide used to make the new solution? (1) 0.60 M     (2) 0.28 M       (3) 0.034 M     (4) 0.068 M      (5) 0.52 M

Answer 1

moles in 280 ml of 2.4 M KBr = 0.24*280/1000= 0.672 moles

KBr----> K+ Br- , 1 mole of KBr contains 1 mole of Br-

hence moles of KbR= 0.672

Let M= Molarity of Calcium Carbinde. moles of Calcium bromide in 120 ml =M*0.12

CaBr2---> Ca+2 + 2Br-

for every mole of CaBr2, there will be 2 mole of Br-, moles of Br- = 2*M*0.12= 0.24M

The total volume, V= 280+120=400ml, its molairty= 1.85M

the total moles= 0.672+0.24M= 1.85*400/1000= 0.74

0.24M= 0.74-0.672=0.068

M= 0.068/0.24=0.283 concentration of CaBr2 ( B is the correct anwer)