Chapter 15.2, Problem 3RQ

Interpretation Introduction

Interpretation: Number of moles of each type of ion in $1.25\text{mol}$ aqueous solution of ${\text{CaCl}}_{\text{2}}$ must be determined.

Concept introduction: Mole is S.I. unit. The number of moles is calculated as ratio of mass of compound to molar mass of compound.

Molar mass is sum of the total mass in grams of all atoms that make up mole of particular molecule that is mass of 1 mole of compound. The S.I unit is $\text{g/mol}$ .

The expression to relate number of moles, mass and molar mass of compound is as follows:

  $\text{Number of moles}=\frac{\text{mass of the compound}}{\text{molar mass of the compound}}$

Answer to Problem 3RQ

  $1.25\text{mol}$ aqueous solution of ${\text{CaCl}}_{\text{2}}$ results into $2.5\text{mol}$ of ${\text{Cl}}^{-}$ and $1.25\text{mol}$ of ${\text{Ca}}^{+2}$ .

Explanation of Solution

For $1.25\text{mol}$ aqueous solution of ${\text{CaCl}}_{\text{2}}$ ,

  ${\text{CaCl}}_{\text{2}}\left(aq\right)\to {\text{Ca}}^{+2}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)$

  ${\text{CaCl}}_{\text{2}}$ dissociate into ${\text{Ca}}^{+2}$ and ${\text{Cl}}^{-}$ ions when dissolved in water. Form equation above it is determined, $1.25\text{mol}$ aqueous solution of ${\text{CaCl}}_{\text{2}}$ results into $1.25\text{mol}\times \text{2 i}\text{.e 2}\text{.5}\text{mol}$ of ${\text{Cl}}^{-}$ and $1.25\text{mol}$ of ${\text{Ca}}^{+2}$ .

Conclusion

  $1.25\text{mol}$ aqueous solution of ${\text{CaCl}}_{\text{2}}$ results into $2.5\text{mol}$ of ${\text{Cl}}^{-}$ and $1.25\text{mol}$ of ${\text{Ca}}^{+2}$ .