Question

1) when you have 2.0M solution of NaCl, what is the molarity of Na+ ion and Cl- ion? 2) If you have 2.0M solution of CaCl2, what is the molarity of Ca2+ and the molarity of Cl-ion in this solution? 3) Comparing molarity of Cl- ion in 2.0M NaCl solution and 2.0M CaCl2 solution, which solution has more Cl- ion? 4) Give formula of the two ions exist in a solution of 0.10M of Al(NO3)3 ? what are their respective ion concentrations?

please answer all.

Answer 1

Question 1)

Molarity of NaCl solution= 2.0 M

Since, 1 mol Na⁺ ions are there in 1 mol of NaCl, molarity of Na⁺ ions = molarity of NaCl solution = 2.0 M

Also, 1 mol of Cl^{​​​​​-}​​​​ ions are there in 1 mol of NaCl, molarity of Cl^{​​​​​-}​​​​ ions = molarity of NaCl solution = 2.0 M

Question 2)

Molarity of CaCl_​​​2​​ solution = 2.0 M

Since, 1 mol of Ca^{​​​​​2+} ions are there in 1 mol of CaCl_​2 solution, molarity of Ca^{​​​​​2+} ions = molarity of CaCl_​​​2​​ solution = 2.0 M

Also, 2 mol of Cl^{​​​​​-}​​​​ ions are there in 1 mol of CaCl_​2 , molarity of Cl^{​​​​​-}​​​​ ions = 2 × molarity of CaCl_​​​2 solution = (2 × 2) M = 4 M.

Question 3

As we calculated in question 1 & 2,

Molarity of Cl^{​​​​​-}​​​​ in CaCl_​​​2 ​​​​​​solution = 4.0 M

Molarity of Cl^{​​​​​-} in NaCl solution = 2.0 M

Also, molarity is directly proportional to no. of moles.

Hence, no. of moles of Cl^{​​​​​-}​​​​ in CaCl_​​​2 solution will be more than that in NaCl solution.

Thus, 2 M CaCl_​​​2 will have more Cl^{​​​​​-} as compared to 2 M NaCl.