Question

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.

Answer 1

`M=2.24M`

Step-by-step 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:

`n_(Cl^-)=3.00L*3.74(molNaCl)/(L)*(1molCl^-)/(1molNaCl) =11.22molCl^-`

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

`M=(11.22mol)/(5.00L)=2.24M`

Best regards!