Question

A solution is 0.1M with respect to KCl and 0.2M with respect to. MgCl2.The molarity of Cl^- ions in the solution is

Answer 1

Answer:

Explanation:

=> First of all lets calculate the total number of CL- ions.

=> One mol of HCL will give one mol of H+ and one mol of CL- when dissociates. Thus, .2 mol gives .2 mol CL-

=> One mol of Bacl2 will give one mole of ba2+ and 2 mol of CL- when dissociates. Hence, 0.1 mol will give .2 mole CL-

=> So the total mols of CL- are .2+.2 = .4

=> Molarity:

molarity M = Total moles of solute/ Volume of solution in liter.

= .4/.5 = 0.8 M

Therefore, the molarity of CL- in solution is 0.8 M.

Explanation:

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Answer 2

Answer:

0.5 M

Explanation:

Given- Concentration of solution with respect to$KCl$ and $MgCl_{2}$.

To find- Molarity of $Cl^{-}$ ions in the solution.

Explanation-

As the given solution has both $KCl$ and $MgCl_{2}$  which are ionic compounds .Ionic compounds are made up of ions and  when they dissolve in water, they dissociate into their respective ions in aqueous form as-

$KCl$→$K^{+} + Cl^{-}$

$MgCl_{2}$→$Mg^{2+} + 2Cl^{-}$

Thus, as per the above equations, 1 M $KCl$ gives 1 M $Cl^{-}$, therefore, 0.1 M $KCl$ produces 0.1 M $Cl^{-}$ .

Similarly, 1 M   $MgCl_{2}$ produces 2 M $Cl^{-}$ ions.

Thus, 0.2 M $MgCl_{2}$ produces $(2$ ×$0.2) = 0.4 M$ $Cl^{-}$ ions.

Thus, the total molarity of $Cl^{-}$ ions can be calculated by adding the concentration of chloride ions obtained from two ionic compounds -$0.1 M + 0.4 M = 0.5 M$.

Hence, the answer is 0.5 M.

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