Question

Calculate the mass percent$(m/m)$of a solution containinglocalid="1654601149382" $26gof{K}_{2}C{O}_3$andlocalid="1654601155761" $724gof{H}_{2}O.$$(9.4)$

Short answer

Hence, the solvent$26g{K}_{2}C{O}_{3}and724g{H}_{2}O$has a mass percent$(m/m)of3.5\%.$

Step-by-step solution

Introduction.

Compounds or solutions are two or more constituents systems that are homogenous. A mixture of treated water and some disintegrating material, such as a solid or gas; nevertheless, solutions of just one solid in an another are also common, such as gold and silver alloys. Any homogenous system with components is referred to as a solution (liquid, solid, or gas).

Step 2:Given data.

The molarity of a solution is described as the amount of solute per litre of solution.

Explanation.

$M$represents it. A solution's molarity is defined as the number of moles of solutes per litre of solution. The following formula describes it:

$Molarity\left(M\right)=\frac{molesossolute}{literofsolution}$

Given data.

The mass is its quantity per unit volume (more accurately, its huxley dense; also known as specific mass).

Explanation.

A product's density is simply the ratio of the its mass to its area. A substance's thickness can be measured as its density of a substance.

$\text{density}\left(\mathrm{g}/{\mathrm{cm}}^3\right)=\frac{\text{mass}\left(\mathrm{g}\right)}{\text{volume}\left({\mathrm{cm}}^3\right)}\text{or}\frac{\text{mass}\left(\mathrm{g}\right)}{\text{volume}\left(\mathrm{mL}\right)}$

Given data.

$\%$ of population:

Mass percentage is a quantitative phrase for a solute that can be estimated of the solution's mass.

Explanation.

The formula:

$masspercentageofsolute=\frac{gramsofsolute}{gramsofsolution}\times 100\%$

Given data.

Use the formula given to get the mass fraction$(m/m)$of a was alllocalid="1654601220309" $26g{K}_{2}C{O}_3$and localid="1654601232885" $724gof{H}_{2}O$:

$\text{Mass percentage of solute}=\frac{\text{grams of solute}}{\text{grams of solution}}\times 100\mathrm{\%}$

Explanation.

$\text{Mass percentage of solute}=\frac{\text{grams of solute}}{\text{grams of solution}}\times 100\mathrm{\%}$

localid="1654601253229" $=\frac{26g{K}_{2}C{O}_3}{26g{K}_{2}C{O}_3+724g{H}_{2}O}\times 100\%$

localid="1654601265721" $=\frac{26g{K}_{2}C{O}_3}{750gsolution}\times 100\%$

$=3.5\%$

The liquid containinglocalid="1654601273605" $26g{K}_{2}C{O}_{3}and724g{H}_{2}O$ has a weight percent$(m/m)of3.5\%$