A 50% sucrose solution lost 255 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 :

Answer: 92.50%

From the given problem, we can say that the total volume of the concentration is:

[tex]255\text{ mL}+300\text{ mL}=555\text{ mL}[/tex]

The initial volume of the solution is 555 mL. After the evaporation, the 50% sucrose solution lost 255 mL.

To find the concentration of the remaining 300 mL, we will use the following formula:

[tex]C_1V_1=C_2V_2[/tex]

Where:

C₁ = initial concentration

C₂ = final concentration

V₁ = initial volume

V₂ = final volume

From the given, we know that:

C₁ = 50% = 0.50

C₂ = ?

V₁ = 555 mL

V₂ = 300 mL

Substitute these to the formula and we will get:

[tex]\begin{gathered} C_{1}V_{1}=C_{2}V_{2} \\ (0.50)(555)=C_2(300) \\ C_2=\frac{(0.50)(555)}{(300)} \\ C_2=0.925=92.50\% \end{gathered}[/tex]

Therefore, we can say that the sucrose concentration in the remaining 300mL is 92.50%

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