Respuesta :
Answer:
[tex]\boxed{\text{1.96 g}}[/tex]
Explanation:
[tex]K = \dfrac{c_{\text{org}}}{c_{\text{aq}}} = 9.73[/tex]
1. First extraction
Let x = mass of compound extracted into ethyl acetate. Then
[tex]\begin{array}{rcl}9.73 & = & \dfrac{x/10.0}{(9.65 - x)/80.0}\\\\9.73\times (9.65 - x)/80.0 & = & x/10.0\\9.73\times (9.65 - x) & = & 8.00x\\93.89 - 9.73x & = & 8.00x\\17.73x & = & 93.89\\x & = & 5.30\\\end{array}[/tex]
So, 5.30 g are extracted into the ethyl acetate.
Mass remaining in water = (9.65 – 5.30) g = 4.35 g
2. Second extraction
[tex]\begin{array}{rcl}9.73 & = & \dfrac{x/10.0}{(4.35 - x)/80.0}\\\\9.73\times (4.35 - x)/80.0 & = & x/10.0\\9.73\times (4.35 - x) & = & 8.00x\\42.33 - 9.73x & = & 8.00x\\17.73x & = & 42.33\\x & = & 2.39\\\end{array}\\\text{So, 2.39 g are extracted into the ethyl acetate.}\\\text{ Mass remaining in water = (4.35 -2.39) g } = \boxed{\textbf{1.96 g}}[/tex]
The partition coefficient is the proportion of the concentration. The mass remaining in the water is 1.96 gm.
What is extraction and partition coefficient?
Given,
- Partition coefficient = 9.73
- Initial Mass of the compound = 9.65 gm
- The volume of water = 80 mL
Calculate the first extraction:
Mass of compound extracted into ethyl acetate = x
Solving for x,
[tex]\begin{aligned}9.73 &= \dfrac{\dfrac{\rm x}{10}}{\dfrac{9.65 - \rm x}{80}}\\\\93.89 - 9.73 \rm \rm x &= 8.00 \rm x\\\\\rm x &= 5.30\end{aligned}[/tex]
Thus, 5.30 g is extracted and the mass remaining in the water is calculated as,
[tex](9.65 - 5.30) \rm g = 4.35 g[/tex]
Calculate the second extraction:
[tex]\begin{aligned}9.73 &= \dfrac{\dfrac{\rm x}{10}}{\dfrac{4.35-\rm x}{80}}\\\\42.33 - 9.73 \rm x &= 8.00 \rm x\\\\\rm x &= 2.39\end{aligned}[/tex]
Thus, 2.39 g is extracted and the mass remaining in the water is calculated as,
[tex](4.35 - 2.39) \rm g = 1.96 g[/tex]
Therefore, 1.96 gm is the mass remaining.
Learn more about partition coefficient here:
https://brainly.com/question/1527402
