Chapter 19, Problem 15STP

Interpretation Introduction

Interpretation:

The chemical which will dissociate into greatest number of particles out of given five is to be calculated.

Concept introduction:

Molarity of the substance helps to calculate the number of moles of that substance. Then number of moles helps to calculate number of particles in aqueous solution.

$Molarity=\frac{numberofmolesofsolute}{volumeofsolutioninliters}$

$numberofparticles=numberofmoles\times 6.023\times {10}^{23}$

Answer to Problem 15STP

Chemical C will dissociate into greatest particles in aqueous solution.

Explanation of Solution

Given data:

$\begin{array}{l}V=500mL=\frac{500}{1000}L\\ Molarity=0.25M\end{array}$

$KCl$ dissociates into $K^{+}andC{l}^{-}$ in aqueous solution.

Therefore number of particles of $\begin{array}{c}KCl=2\times 6.023\times {10}^{23}\times 0.250\times 0.5\\ =1.51\times {10}^{23}\end{array}$

Number of moles of $\begin{array}{c}C{H}_{3}OH=6.023\times {10}^{23}\times 0.250\times 0.5\\ =0.75\times {10}^{23}\end{array}$

$Ba{\left(OH\right)}_2$ dissociates into three ions.

Therefore, number of moles of $\begin{array}{c}Ba{\left(OH\right)}_2=3\times 6.023\times {10}^{23}\times 0.250\times 0.5\\ =2.26\times {10}^{23}\end{array}$

Number of moles of $\begin{array}{c}C{H}_{3}COOH=6.023\times {10}^{23}\times 0.250\times 0.5\\ =0.75\times {10}^{23}\end{array}$

$NaOH$ dissociate into two ions.

Therefore number of moles of $\begin{array}{c}NaOH=2\times 6.023\times {10}^{23}\times 0.250\times 0.5\\ =1.51\times {10}^{23}\end{array}$

$Ba{\left(OH\right)}_2$ has greatest number of particles in aqueous solution.

Conclusion

Because $Ba{\left(OH\right)}_2$ dissociates into maximum ions in aqueous solution, it has greatest number of particles.