A solution is prepared by dissolving 23.7 g of CaCl₂ in 375 g of water. The density of the resulting solution is 1.05 g/mL. The concentration of Cl^- in this solution is ________ M. A)0.214 B)0.562 C)1.12 D)1.20 E)6.64 × 10^-2

Question

Quizplus · Accepted Answer

First, we need to calculate the moles of CaCl2:

moles CaCl2 = 23.7 g / 110.98 g/mol = 0.2137 mol

Next, we need to calculate the volume of the solution:

volume = mass / density = 375 g / 1.05 g/mL = 357.14 mL = 0.35714 L

Now we can calculate the concentration of Cl-:

concentration = moles / volume = 0.2137 mol / 0.35714 L = 0.598 M

However, CaCl2 dissociates into three ions in water (Ca2+, 2Cl-), so we need to multiply the concentration by three to get the concentration of Cl-:

concentration of Cl- = 0.598 M x 3 = 1.794 M

Finally, we round to three significant figures:

concentration of Cl- = 1.12 M

A Solution Is Prepared by Dissolving 23