Question

For $$20ml$$ of $$0.2m$$$$A{l_2}{\left( {S{O_4}} \right)_3}$$ is mixed with $$20ml$$ of $$6.6m$$ $$BaC{l_2}$$ the concentration of $$C{l^ - }$$ ion in solution is:
A. $0.2$
B. $6.6$
C. $0.02$
D. $0.06$

Answer 1

Hint: In the above question, solutions of Aluminium sulphate $$(A{l_2}{\left( {S{O_4}} \right)_3})$$ and Barium chloride $$(BaC{l_2})$$ are mixed together. We are given the volumes and concentrations of both these solutions. Therefore, in order to find the total concentration of chloride ions $$(C{l^ - })$$ in the mixture solution, we will first calculate the total moles of chloride ions from the barium sulfate solution and then using the formula of molarity, we will calculate the new concentration of chloride.

Complete answer:
We are given a mixture of $$20ml$$ of $$0.2m$$$$A{l_2}{\left( {S{O_4}} \right)_3}$$ and $$20ml$$ of $$6.6m$$ $$BaC{l_2}$$ . We get the chloride ions from Barium chloride only. So, we will first calculate the number of moles of chloride ions coming from the $$20ml$$ of $$6.6m$$ Barium chloride solution as follows:
Total amount of chloride ions in the solution (n) is equal to the concentration of chloride ions (M) multiplied by the volume of the solution (V).
$n = M \times V\_\_\_\_(1)$
One mole of Barium chloride produces two moles of chloride ions, so, the concentration of chloride ions is:
$ = 2 \times 6.6$
$ = 13.2M$
Put these values in the above mentioned formula:
$n = \dfrac{{20}}{{1000}} \times 13.2$
We have divided volume by $1000$ because volume should be in Liter.
Total volume of the mixture is:
$20 + 20 = 40ml$
Concentration of chloride ion is denoted as M.
$M = \dfrac{{n \times 1000}}{{V(ml)}}$
Put the above calculated value of n in this formula:
$M = \dfrac{{20 \times 13.2}}{{1000}} \times \dfrac{{1000}}{{40}}$
On solving this, we get:
$M = 6.6M$
Therefore, the correct option is B. $6.6.$

Note:
We should remember that in order to solve these types of questions we should know about the term number of moles, molarity and molality. Molarity is defined as the number of moles of solute present in one litre of the solution while molality is defined as the number of moles of solute present in one kilogram of the solvent.