91Ó°ÊÓ

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.

The following data is displayed in the problem:

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

Determine the mass proportion using the continuity formula:

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

Rearrange the computation so that the values are substituted,

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

The following data is displayed in the problem:

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

Determine the mass proportion using the continuity formula:

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

Rearrange the computation so that the values are substituted,

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

The following data is displayed in the problem:

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

Determine the mass proportion using the continuity formula:

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

Rearrange the computation so that the values are substituted,

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