Chapter 16.2, Problem 13SP

Interpretation Introduction

Interpretation: To calculate the moles of solute present in $250ml$ of $2.0M$

  ${\text{CaCl}}_{\text{2}}$ and total grams of ${\text{CaCl}}_{\text{2}}$ .

Concept introduction: Molarity is defined as the ratio of moles of solute to that of the volume of solvent that is used to dissolve the given solute.

Answer to Problem 13SP

The moles of solute present in the solution is $0.50mol$ and $55g$ of ${\text{CaCl}}_{\text{2}}$ .

Explanation of Solution

Moles of ${\text{CaCl}}_{\text{2}}$

In $1L$ solution the number of moles $=2.0\text{ mol}$

So, in $250ml$ solution :

  $\begin{array}{l}\text{number of moles}=0.250Lsolution\times \frac{2.0molCaC{l}_2}{1Lsolution}\\ =0.50{\text{ mol CaCl}}_{\text{2}}\end{array}$

Mass of $CaC{l}_2$

1mol of $CaC{l}_2$ contain mass $=(40.08+2\times 35.45)g$

  $=110.98g$

So, $0.50mol$ of $CaC{l}_2$ contain mass $=0.50molCaC{l}_2\times \frac{110.98gCaC{l}_2}{1molCaC{l}_2}$

  $=55g$

Conclusion

The moles of solute present in the solution is $0.50mol$ and $55g$ of ${\text{CaCl}}_{\text{2}}$ .