Chapter 6, Problem 6.42QP

(a)

Interpretation Introduction

Interpretation:

The amount of solute needed to prepare $250gof5\%\left(m/m\right)$ ammonium 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 6.42QP

The amount of solute needed to prepare $250gof5\%\left(m/m\right)$ ammonium chloride solution is  $12.5g.$

Explanation of Solution

The amount of solute needed to prepare $250gof5\%\left(m/m\right)$ ammonium chloride solution can be calculated as follows,

    $\begin{array}{c}\%\left(\frac{m}{m}\right)=\frac{gofsolute}{gofsolution}\times 100\%\\ \\ amountofsolute=\frac{\left(\%\left(\frac{m}{m}\right)\right)\times gofsolution}{100\%}\\ \\ =\frac{\left(5\%\left(m/m\right)\right)\times 250g}{100\%}\\ \\ =12.5g.\end{array}$

Amount solute required is $12.5g.$

(b)

Interpretation Introduction

Interpretation:

The amount of solute needed to prepare $250gof3.50\%\left(m/m\right)N{a}_{2}C{O}_3$ solution has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 6.42QP

The amount of solute needed to prepare $250gof3.50\%\left(m/m\right)N{a}_{2}C{O}_3$ solution is $8.75g.$

Explanation of Solution

The amount of solute needed to prepare $250gof3.50\%\left(m/m\right)N{a}_{2}C{O}_3$ solution can be calculated as follows,

    $\begin{array}{c}\%\left(\frac{m}{m}\right)=\frac{gofsolute}{gofsolution}\times 100\%\\ \\ amountofsolute=\frac{\left(\%\left(\frac{m}{m}\right)\right)\times gofsolution}{100\%}\\ \\ =\frac{\left(3.50\%\left(m/m\right)\right)\times 250g}{100\%}\\ \\ =8.75g.\end{array}$

Amount solute required is $8.75g.$