Chapter 16, Problem 61A

(a)

Interpretation Introduction

Interpretation: The amount of solute in grams required to prepare the solution of $2500\text{ g}$ of saline solution is to be calculated.

Concept Introduction: Mass percent is the grams of solute present in 100g of the solution. It is represented as $(m/m)$ % or $(w/w)$ %.

Answer to Problem 61A

The mass of $\text{NaCl}$ required is $23\text{ g}$ .

Explanation of Solution

Given, The value $0.90\%$ of $\text{NaCl}$ solution $(m/m)$ means $0.90\text{ g}$ of $\text{NaCl}$ in $100\text{ g}$ of the solvent.

So,

The mass of $\text{NaCl}$ required to make $2500\text{ g}$ of saline solution is calculated as follows:

  $\begin{array}{c}mass\text{of NaCl}=\frac{2500\text{ g of solution}\times 0.90\text{ of NaCl}}{100\text{ of solution}}\\ =22.5\text{ g}\\ \approx \text{23 g}\end{array}$

The mass of $\text{NaCl}$ required is $23\text{ g}$ .

(b)

Interpretation Introduction

Interpretation: The amount of solute in grams required to prepare the solution of $0.050\text{ kg}$ of $4.0\%$ of ${\text{MgCl}}_2$ is to be calculated.

Concept Introduction: Mass percent is the grams of solute present in 100g of the solution. It is represented as $(m/m)$ % or $(w/w)$ %.

Answer to Problem 61A

The mass of ${\text{MgCl}}_2$ required is $\text{2}\text{.0 g}$ .

Explanation of Solution

Given,

The value $4.0\%$ of ${\text{MgCl}}_2$ solution $(m/m)$ means $4.0\text{ g}$ of ${\text{MgCl}}_2$ in $100\text{ g}$ of the solvent.

Mass of the solution is $0.050\text{ kg}$ .

Convert kilogram into grams.

  $\begin{array}{c}0.050\text{ kg}=0.050\text{ kg}\times \frac{1000\text{ g}}{1\text{ kg}}\\ =50\text{ g}\end{array}$

So,

The mass of $\text{NaCl}$ required to make $50\text{ g}$ of saline solution is calculated as follows:

  $\begin{array}{c}mass\text{of NaCl}=\frac{50\text{ g of solution}\times 4.0{\text{ of MgCl}}_2}{100\text{ of solution}}\\ =\text{2}\text{.0 g}\end{array}$

The mass of ${\text{MgCl}}_2$ required is $\text{2}\text{.0 g}$ .