300 ml of NaCl of 3-0M is added to 200 ml of BaCl2 nbsp-of molarity 4-0 M- Claculate the concentration of Cl ion in the resulting solution-

Start by getting the moles of Cl in each, nbsp; No of moles of NaCl = molarity #215;volume = 3 #215;0.300 = 0.90 mol NaCl In Nacl there is 1 mol of Cl ...

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Standard XII

Chemistry

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300 ml of NaC...

Question

300 ml of NaCl of 3.0M is added to 200 ml of BaCl₂of molarity 4.0 M. Claculate the concentration of Cl ion in the resulting solution.

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Solution

Start by getting the moles of Cl in each,

No of moles of NaCl = molarity $\times$ volume = 3 $\times$ 0.300 = 0.90 mol NaCl
In Nacl there is 1 mol of Cl
Therefore 0.90 mol Cl^-

No of moles of BaCl₂ = molarity $\times$ volume = 4 $\times$ 0.200 = 0.80 mol BaCl₂
In BaCl₂ there is 2 mol of Cl^-
Therefore 1.6 mol Cl^-

Now add both amounts, 0.9 mol Cl^- + 1.6 mol Cl^- = 2.5 mol Cl^-

Finally calculate the molarity

2.5 mol Cl^-/0.5 l = 5 M

or 5 M of Cl^-

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300 ml of NaCl of 3.0M is added to 200 ml of BaCl2 of molarity 4.0 M. Claculate the concentration of Cl ion in the resulting solution.

300 ml of NaCl of 3.0M is added to 200 ml of BaCl₂of molarity 4.0 M. Claculate the concentration of Cl ion in the resulting solution.