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You have 3.00 L of a 3.74 M solution of NaCl(aq) called solution A. You also have 2.00 L of a 2.00 M solution of AgNO3(aq) called solution B. You mix these solutions together to make solution C. Calculate the concentration (in M) of Cl ions in solution C.

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Answer:

[tex]M=2.24M[/tex]

Explanation:

Hello!

In this case, since the molarity and volume of the solutions are given, we are able to compute the moles of chloride ions as they are present in NaCl only:

[tex]n_{Cl^-}=3.00L*3.74\frac{molNaCl}{L}*\frac{1molCl^-}{1molNaCl} =11.22molCl^-[/tex]

Next, since the mixing of solution A and B lead to a volume of 5.00 L, the concentration of chloride ions turns out:

[tex]M=\frac{11.22mol}{5.00L}=2.24M[/tex]

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