What volume of 0.8m alcl3 solution should be mixed with 50 ml of 0.2m cacl2 solution to get solution of chloride ion concentration equal to 0.6m?
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let V is the required volume.
number of moles of AlCl3 = Vx0.8
number of moles of Cl- in AlCl3 (n1) = 3 x V x 0.8
number of moles of Cl- in CaCl2 (n2) = (2 x 50 x 0.2)/ 1000
for the solution..
0.6 = (n1+n2)/(v + 50/1000)
by substituting the values and solving above equation...
=> (2.4-1.6)v= 0.02-0.03
=> 1.8v = 0.01
=> v = 5.55 × 10^-³ L = 5.55 ml
number of moles of AlCl3 = Vx0.8
number of moles of Cl- in AlCl3 (n1) = 3 x V x 0.8
number of moles of Cl- in CaCl2 (n2) = (2 x 50 x 0.2)/ 1000
for the solution..
0.6 = (n1+n2)/(v + 50/1000)
by substituting the values and solving above equation...
=> (2.4-1.6)v= 0.02-0.03
=> 1.8v = 0.01
=> v = 5.55 × 10^-³ L = 5.55 ml
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