Question

In the laboratory you dissolve 19.8 g of calcium bromide in a volumetric flask and add water to a total volume of 500 ml. M. What is the molarity of the solution? What is the concentration of the calcium cation? м. What is the concentration of the bromide anion?

Answer 1

mass of CaBr2 taken = 19.8 gram , molar mass of CaBr2 = 199.89 gram/mol

no of moles = mass in gram/molar mass substitute values in this formula to find no of moles of CaBr2 used

no of moles of CaBr2 taken = 19.8/199.89 = 0.099 moles CaBr2 taken

volume of solution = 500 ml convert this volume in liter by using following formula

volume in liter = volume in ml/1000 substitute values in this formula

volume of solution in liter = 500/1000 = 0.5 liter

molarity = no of moles/volume in liter , substitute above calculated values in this formula to find molarity

molarity of solution = 0.099/0.5 = 0.198 M

ans molarity of solution is 0.198 molar

CaBr2 is strong electrolyte which dissociate completly in water as given in following reaction

CaBr2 + H2O  $\rightarrow$ Ca²⁺_(aq) + 2Br^-_(aq)

according to dissociation reaction of CaBr2 1 mole of CaBr2 produces 1 mole of Ca²⁺ therefore molarity of Ca²⁺ is same as molarity of CaBr2  

therefore molarity of Ca²⁺ = molarity of CaBr2 = 0.198 M

ans concentration of calcium cation is = 0.198 M

CaBr2 is strong electrolyte which dissociate completly in water as given in following reaction

CaBr2 + H2O  $\rightarrow$ Ca²⁺_(aq) + 2Br^-_(aq)

according to dissociation reaction of CaBr2 1 mole of CaBr2 produces 2 moles of Br^- therefore molarity of Br^- is double than molarity of CaBr2  

therefore molarity of Br^- = 2 x molarity of CaBr2 = 0.198 x 2 = 0.396 M

ans concentration of bromide anion is = 0.396M