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Sagot :
The reciprocal of the density, ρ, of a mixture is equal to the sum of the reciprocal of the densities of the constituents multiplied by their respective proportions ([tex]x_i[/tex]) in the solution, [tex]\displaystyle \dfrac{1}{\rho_{(mixture)}} = \sum \dfrac{x_i}{\rho _i}[/tex]
The percentage of water in the 2.75 g/mL SPT solution is [tex]\underline{16. \bar6 \ \%}[/tex]
Reason:
Known parameter:
Density of SPT = 3.10 g/mL
Density of water = 1.00 g/mL
Required:
The percentage of water in 2.75 g/mL SPT solution
Solution:
Let x the volume of water added per 100 mL of 2.75 g/mL SPT solution, we have;
Mass of water in the solution = x mL × 1.00 g/mL = x g
Mass of commercial SPT = (100 - x) mL × 3.10 g/mL = (310 - 3.1·x) g
[tex]Density = \dfrac{Mass}{Volume}[/tex]
The density of the 2.75 g/mL SPT solution is given as follows;
[tex]Density \ of \ 100\ mL, \ 2.75 \ g/mL = \dfrac{(310 - 3.1\cdot x + x) \ g}{100 \ mL}[/tex]
100 mL × 2.75 g/mL =( 310 - 2.1·x) g
310 - 275 = 2.1·x
[tex]x = \dfrac{310 - 275}{2.1} = \dfrac{50}{3} = 16.\overline 6[/tex]
Given that the reference quantity of the proportion of water in the solution is 100 mL of the 2.75 g/mL SPT solution, the percentage (per 100) of water in the overall solution is the value of x
Therefore;
- The percent of the SPT solution that must consist of water to give an overall density of 2.75 g/mL is x = [tex]\displaystyle {\underline {16.\overline 6 \ \%} }[/tex]
Learn more here: https://brainly.com/question/13346179
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