Q. 9.42 - An Introduction to General, Organic, and Biological Chemistry

Question

Calculate the grams or milliliters of solute needed to prepare the following:

a.$\text{ 150 }\mathrm{g}\text{ of a }40.0\mathrm{\%} (\mathrm{m}/\mathrm{m}) {\mathrm{LiNO}}_3$ solution.

b. $450 \mathrm{mL}\text{ of a }2.0\mathrm{\%} (\mathrm{m}/\mathrm{v}) \mathrm{KOH}$ solution.

c. $225 \mathrm{mL}\text{ of a }15\mathrm{\%} (\mathrm{v}/\mathrm{v})$ isopropyl alcohol solution.

Step-by-Step Solution

Answer

(part a) Weight of $\mathrm{150 }\mathrm{g}\text{ of a }40.0\mathrm{\%} (\mathrm{m}/\mathrm{m}) {\mathrm{LiNO}}_3$ solution is $60 g.$

(part b) Weight of $450 \mathrm{mL}\text{ of a }2.0\mathrm{\%} (\mathrm{m}/\mathrm{v}) \mathrm{KOH}$ solution is $9 g.$

(part c) Weight of $225 \mathrm{mL}\text{ of a }15\mathrm{\%} (\mathrm{v}/\mathrm{v})$ isopropyl alcohol solution is $33.75 mL.$

Step1: Introduction (part a).

Solute concentration is explained in terms of volume or molarity. Every single cent equation to determine a compound's molarity and also the quantity of every element inside one molar ratio.

Step2: Given Information (part a).

The following data is displayed in the problem:

$\mathrm{150 }\mathrm{g}\text{ of a }40.0\mathrm{\%} (\mathrm{m}/\mathrm{m}) {\mathrm{LiNO}}_3$ solution.

Step3: Explanation (part a).

Determine the mass proportion using the continuity formula:

$\text{ Mass percent }(\mathrm{m}/\mathrm{m})=\frac{\text{ Solute mass }}{\text{ Solution mass }}\times 100\mathrm{\%}$

Rearrange the computation so that the values are substituted,

$$\begin{gathered} \text{ Solute mass }=\frac{\text{ Mass percent }(\mathrm{m}/\mathrm{m})\times \text{ Solution mass }}{100\mathrm{\%}} \\ =\frac{40.0\mathrm{\%}\times 150\mathrm{g}}{100\mathrm{\%}} \\ =60 g \end{gathered}$$

Step4: Given Information (part b).

The following data is displayed in the problem:

$450 \mathrm{mL}\text{ of a }2.0\mathrm{\%} (\mathrm{m}/\mathrm{v}) \mathrm{KOH}$ solution.

Step5: Explanation (part b).

Determine the mass proportion using the continuity formula:

$\text{ Mass percent }(\mathrm{m}/\mathrm{m})=\frac{\text{ Solute mass }}{\text{ Solution mass }}\times 100\mathrm{\%}$

Rearrange the computation so that the values are substituted,

$$\begin{gathered} \text{ Solute mass }=\frac{\text{ Mass percent }(\mathrm{m}/\mathrm{m})\times \text{ Solution mass }}{100\mathrm{\%}} \\ =\frac{2.0\mathrm{\%}\times 450 mL}{100\mathrm{\%}} \\ =9 g \end{gathered}$$

Step6: Given Information (part c).

The following data is displayed in the problem:

$225 \mathrm{mL}\text{ of a }15\mathrm{\%} (\mathrm{v}/\mathrm{v})$ isopropyl alcohol solution.

Step7: Explanation (part c).

Determine the mass proportion using the continuity formula:

$\text{ Mass percent }(\mathrm{m}/\mathrm{m})=\frac{\text{ Solute mass }}{\text{ Solution mass }}\times 100\mathrm{\%}$

Rearrange the computation so that the values are substituted,

$$\begin{gathered} \text{ Solute mass }=\frac{\text{ Mass percent }(\mathrm{m}/\mathrm{m})\times \text{ Solution mass }}{100\mathrm{\%}} \\ =\frac{15.0\mathrm{\%}\times 225 mL}{100\mathrm{\%}} \\ =33.75 mL \end{gathered}$$