91Ó°ÊÓ

Given:

$25gofKCl$

$125gof{H}_{2}O$

The term "mass percent" refers to the number of grammes of solute dissolved in $100g$of solution.

$$\begin{gathered} \text{Mass of solution}\left(\mathrm{g}\right)=\text{Mass of solute}\left(\mathrm{g}\right)+\text{Mass of solvent}\left(\mathrm{g}\right) \\ =25g+125g \\ =150g \end{gathered}$$

Mass percent equation is,

$\text{Mass percent}(\mathrm{m}/\mathrm{m})=\frac{\text{Mass of solute}\left(\mathrm{g}\right)}{\text{Mass of solution}\left(\mathrm{g}\right)}×100\mathrm{\%}\text{.}$

Substitute the values,

$$\begin{gathered} \text{Mass percent}(\mathrm{m}/\mathrm{m})=\frac{25\mathrm{gKCl}}{150\mathrm{g}}×100\mathrm{\%} \\ =17\%(\mathrm{m}/\mathrm{m})\mathrm{KCl} \end{gathered}$$

Given:

$12g$of sucrose.

$225g$of tea solution.

Mass percent equation is,

$\mathrm{Mass}\mathrm{percent}(\mathrm{m}/\mathrm{m})=\frac{\text{Mass of solute}\left(\mathrm{g}\right)}{\text{Mass of solution}\left(\mathrm{g}\right)}×100\mathrm{\%}\text{.}$

Substitute the values,

$$\begin{gathered} \text{Mass percent}(\mathrm{m}/\mathrm{m})=\frac{8.0\mathrm{g}{\mathrm{CaCl}}_2}{80.0g\mathrm{of}{\mathrm{CaCO}}_3}×100\mathrm{\%} \\ =10\%(\mathrm{m}/\mathrm{m}){\mathrm{CaCl}}_2 \end{gathered}$$

Given:

$8.0gofCaC{l}_2$

$80.0gofCaC{O}_2$solution.

The mass percent equation is,

$\mathrm{Mass}\mathrm{percent}(\mathrm{m}/\mathrm{m})=\frac{\text{Mass of solute}\left(\mathrm{g}\right)}{\text{Mass of solution}\left(\mathrm{g}\right)}×100\mathrm{\%}\text{.}$

Substitute the values,

$$\begin{gathered} \text{Mass percent}(\mathrm{m}/\mathrm{m})=\frac{12g\text{sucrose}}{225\mathrm{g}}×100\mathrm{\%} \\ =5.3\%(\mathrm{m}/\mathrm{m})\mathrm{sucrose} \end{gathered}$$