A 45% sucrose solution lost 260 mL by evaporation. After evaporation what is the sucrose concentration in the remaining 300 mL? Round your final answer to 2 decimal places if necessary.

Respuesta :

84%

Explanation

the concentration is given by

[tex]\text{concentration = }\frac{volume\text{ of sucrose}}{\text{total volume}}\cdot100\text{ \%}[/tex]

hence

Step 1

for the initial solution

[tex]45\text{ \%=}\frac{volumesucre}{\text{total volume}}\cdot100\rightarrow equation(1)[/tex]

Step 2

after the evaporation

[tex]\begin{gathered} x\text{ \%=}\frac{volumesucrose}{(initial\text{ volume-evaporatio)}} \\ x\text{ \%=}\frac{volumesucrose}{(initial\text{ volume-260)}=300} \\ x\text{ \%=}\frac{volumesucrose}{300}\rightarrow equation(2) \end{gathered}[/tex]

[tex]\begin{gathered} (initial\text{ volume-260)}=300 \\ add\text{ 260 in both sides} \\ (initial\text{ volume-260)+260}=300+260 \\ initial\text{ volume=560} \end{gathered}[/tex]

now, replace this value in step 1 to find the initial volume of sucrose

[tex]\begin{gathered} 45\text{ \%=}\frac{volumesucre}{\text{5}60} \\ 0.45\cdot560mL=\text{volume sucrose} \\ 252\text{ mL = Volume of sucrose} \end{gathered}[/tex]

now, replace this value in eq(2)

[tex]\begin{gathered} x\text{ \%=}\frac{volumesucrose}{300}\rightarrow equation(2) \\ x\text{ \%=}\frac{252}{300}\cdot100 \\ x\text{ \%=84 \%} \end{gathered}[/tex]

therefore, the answer is

84%

I hope this helps you

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