Chapter 15, Problem 16A

Interpretation Introduction

Interpretation:

Mass of NaCl and Na₂CO₃ present in 285 gm of a solution have to be calculated.

Concept Introduction:

Composition of a solution can be defined by expressing its concentration. The concentrations of solutions can be expressed in different ways, which are involved in the quantity of solute and the quantity of solution or solvent. Various methods are used to describe the concentration of the solution quantitatively. Some commonly used quantitative concentration terms are percent by mass, percent by volume, molarity, molality and mole fraction.

Mass percentage or percent by mass (w/W %): The mass percentage of a component of a solution is defined as:

  $\mathrm{Mass}\%ofacomponent=\frac{Mass of the component in the solution}{Total mass of the solution}\times 100$

Answer to Problem 16A

14.25 gm of NaCl and 21.38 gm of Na₂CO₃ are present in 285 gm of a solution.

  $\mathrm{Mass}\%ofacomponent=\frac{Mass of the component in the solution}{Total mass of the solution}\times 100$

Data given: Total mass of the solution = 285 gm

Percent by mass NaCl = 5.00%

  $\begin{array}{l}PercentbymassNaCl=\frac{Massof NaCl}{Total mass of NaClsolution}\times 100\\ Massof NaCl=\frac{PercentbymassNaCl\times Total mass of NaClsolution}{100}\\ Massof NaCl=\frac{5.00\times 285gm}{100}=14.25gm\end{array}$

Percent by mass Na₂CO₃ = 7.50%

  $\begin{array}{l}PercentbymassN{a}_{2}C{O}_3=\frac{Massof N{a}_{2}C{O}_3}{Total mass of N{a}_{2}C{O}_{3}solution}\times 100\\ Massof N{a}_{2}C{O}_3=\frac{PercentbymassN{a}_{2}C{O}_3\times Total mass of N{a}_{2}C{O}_{3}solution}{100}\\ Massof N{a}_{2}C{O}_3=\frac{7.50\times 285gm}{100}=21.38gm\end{array}$

Explanation of Solution

  $\mathrm{Mass}\%ofacomponent=\frac{Mass of the component in the solution}{Total mass of the solution}\times 100$

Data given: Total mass of the solution = 285 gm

Percent by mass NaCl = 5.00%

  $\begin{array}{l}PercentbymassNaCl=\frac{Massof NaCl}{Total mass of NaClsolution}\times 100\\ Massof NaCl=\frac{PercentbymassNaCl\times Total mass of NaClsolution}{100}\\ Massof NaCl=\frac{5.00\times 285gm}{100}=14.25gm\end{array}$

Percent by mass Na₂CO₃ = 7.50%

  $\begin{array}{l}PercentbymassN{a}_{2}C{O}_3=\frac{Massof N{a}_{2}C{O}_3}{Total mass of N{a}_{2}C{O}_{3}solution}\times 100\\ Massof N{a}_{2}C{O}_3=\frac{PercentbymassN{a}_{2}C{O}_3\times Total mass of N{a}_{2}C{O}_{3}solution}{100}\\ Massof N{a}_{2}C{O}_3=\frac{7.50\times 285gm}{100}=21.38gm\end{array}$