Q. 9.94 - An Introduction to General, Organic, and Biological Chemistry

Question

Calculate the mass percent  $(m/m)$ of a solution containing  $26 g of K_{2}C{O}_3$ and $724 g of H_{2}O.$$(9.4)$

Step-by-Step Solution

Answer

Hence, the solvent  $26 g K_{2}C{O}_3 and 724 g H_{2}O$ has a mass percent  $(m/m) of 3.5\%.$

Step 1: 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.

Step 3: 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:

$Molarity \left(M\right)=\frac{moles os solute}{liter of solution}$

Step 5: Given data.

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

Step 5: 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.

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

Step 6: Given data.

$\%$ of population:

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

Step 7: Explanation.

The formula:

$mass percentage of solute=\frac{grams of solute}{grams of solution} \times 100\%$

Step 8: Given data.

Use the formula given to get the mass fraction  $(m/m)$ of a was all  $26 g K_{2}C{O}_3$ and $724 g of H_{2}O$:

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

Step 9: Explanation.

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

$=\frac{26 g K_{2}C{O}_3}{26 g K_{2}C{O}_3 + 724 g H_{2}O}\times 100\%$

$=\frac{26 g K_{2}C{O}_3}{750 g solution}\times 100\%$

$=3.5\%$

The liquid containing  $26 g K_{2}C{O}_3 and 724 g H_{2}O$  has a weight percent  $(m/m) of 3.5\%$