Chapter 6, Problem 6.41QP

(a)

Interpretation Introduction

Interpretation:

The amount of solute needed to prepare $250gof0.9\%\left(m/m\right)$ sodium chloride solution has to be calculated.

Concept Introduction:

The mass $/$ mass percent can be expressed as follows,

    $\%\left(\frac{m}{m}\right)=\frac{gofsolute}{gofsolution}\times 100\%$

Answer to Problem 6.41QP

The amount of solute needed to prepare $250gof0.9\%\left(m/m\right)$ sodium chloride solution is  $2.25g.$

Explanation of Solution

The amount of solute needed to prepare $250gof0.9\%\left(m/m\right)$ sodium chloride solution can be calculated as follows,

    $\begin{array}{c}\%\left(\frac{m}{m}\right)=\frac{gofsolute}{gofsolution}\times 100\%\\ \\ amountofsolute=\frac{\left(\%\left(\frac{m}{m}\right)\right)\times gofsolution}{100\%}\\ \\ =\frac{\left(0.9\%\left(m/m\right)\right)\times 250g}{100\%}\\ \\ =2.25g.\end{array}$

Amount solute required is $2.25g.$

(b)

Interpretation Introduction

Interpretation:

The amount of solute needed to prepare $250gof1.25\%\left(m/m\right)$ sodium acetate solution has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

The amount of solute needed to prepare $250gof1.25\%\left(m/m\right)$ sodium acetate solution can be calculated as follows,

    $\begin{array}{c}\%\left(\frac{m}{m}\right)=\frac{gofsolute}{gofsolution}\times 100\%\\ \\ amountofsolute=\frac{\left(\%\left(\frac{m}{m}\right)\right)\times gofsolution}{100\%}\\ \\ =\frac{\left(1.25\%\left(m/m\right)\right)\times 250g}{100\%}\\ \\ =3.125g.\end{array}$

Amount solute required is $3.125g.$