Chapter 15, Problem 61A

Interpretation Introduction

Interpretation: The solution that has the highest concentration needs to be determined.

Concept introduction: Molarity can be defined as the number of moles of solute in 1 litre of solution. Thus,

  $\text{Molarity(}M\text{)}=\frac{\text{Number of moles of solute}}{\text{Volume of the solution in litre}}$

So greater the molarity more concentrated is the solution.

Here for all solutions volume is same and hence, $\text{molarity}\propto \text{number of moles}$ .

Answer to Problem 61A

Solution 2 is most concentrated.

Explanation of Solution

  $\begin{array}{c}\text{Number of moles of sodium sulfate (}N{a}_{2}S{O}_4\text{) =}\frac{\text{mass of }N{a}_{2}S{O}_4}{\text{molar mass of }N{a}_{2}S{O}_4}\\ =\frac{100.0g}{\text{142}\text{.04 }g/mol}\\ =0.704mol\end{array}$

Hence $\text{sodium sulfate (}N{a}_{2}S{O}_4\text{) }$ solution is most concentrated.

Conclusion

  $\begin{array}{c}\text{Number of moles of cesium chloride (}CsCl\text{) = }\frac{\text{mass of }CsCl}{\text{molar mass of }CsCl}\\ =\frac{100.0g}{\text{168}\text{.36 }g/mol}\\ =0.593mol\end{array}$

  $\begin{array}{c}\text{Number of moles of cesium fluoride (CsF) =}\frac{\text{mass of CsF}}{\text{molar mass of CsF }}\\ =\frac{100.0g}{151.9g/mol}\\ =0.658mol\end{array}$

  $\begin{array}{c}\text{Number of moles of sodium phosphate (}N{a}_{3}P{O}_4\text{) =}\frac{\text{mass of }N{a}_{3}P{O}_4}{\text{molar mass of }N{a}_{3}P{O}_4}\\ =\frac{100.0g}{\text{163}\text{.94 }g/mol}\\ =0.609mol\end{array}$

Hence, other options cannot be correct.