1.What Is The Percent By Volume Of A Solution Formed By Mixing 35ml Of Isopropanol With 55ml Water?,

1.what is the percent by volume of a solution formed by mixing 35ml of isopropanol with 55ml water?

 

Answer:

Percentage by Volume

Problem #1

Given:

volume of solute (isopropanol) = 25 mL

volume of solvent (water) = 45 mL

Asked: % volume

Equation: % volume = \frac{volume of solute}{volume of solution}

volumeofsolution

volumeofsolute

x 100%

Solution:

% volume = \frac{volume of solute}{volume of solution}

volumeofsolution

volumeofsolute

x 100%

= \frac{volume of solute}{volume of solute +solvent}

volumeofsolute+solvent

volumeofsolute

x 100%

= \frac{25 mL}{25 mL + 45mL}

25mL+45mL

25mL

x 100%

= \frac{25 mL}{70 mL}

70mL

25mL

x 100%

= 0.3571 x 100%

= 35.71 %

Problem #2

Given:

volume of solute (acetic acid) = 160 mL

volume of solution (stop bath) = 650 mL

Asked: % volume

Equation: % volume = \frac{volume of solute}{volume of solution}

volumeofsolution

volumeofsolute

x 100%

Solution:

% volume = \frac{volume of solute}{volume of solution}

volumeofsolution

volumeofsolute

x 100%

= \frac{160 mL}{650mL}

650mL

160mL

x 100%

= 0.2462 x 100%

= 24.62 %

Explanation:

Solutions are homogeneous mixtures that are made of solute dissolved in a solvent. The amount of solute in the solution tells us of the concentration of the solution. This can be expressed either as percent mass, percent volume or percent mass/volume. If the solute and solvent are in liquid or gas, percent volume may be used. The volume percent of a solution is defined as the ratio of the volume of solute that is present in a solution, relative to the volume of the solution, as a whole. This ratio then is multiplied by 100 to get the percent volume:

% volume = \frac{volume of solute}{volume of solution}

volumeofsolution

volumeofsolute

x 100%

Since a solution is a combination of a solute and a solvent, the volume of a solution, as a whole, is equal to the sum of the volumes of the solute and the solvent that it contains. Therefore, we can use the following equation to