The distribution coefficient between methylene chloride and water for solute Y is 14. An amount of 72.0 g of Y is dissolved in 180 mL of water. a) What weight of Y would be removed from water with a single extraction with 180-mL of methylene chloride? Report to 1 decimal place. b) What total weight of Y would be removed from water (the original solution) with two successive extractions with 90-mL portions each of methylene chloride?

a)Weight of Y extracted from water in single extraction = 67.2 g

b) Total weight of Y extracted from water in two successive extractions = 70.9 g

Explanation:

Distribution coefficient, Kd = Cs / Cm

where Cs  is concentration of solute in methylene chloride; Cm is concentration of solute in water

a) Let x g of solute Y be extracted into methylene chloride

72.0 - x g of Y will be left in water

Kd = (x/180 mL of methylene chloride) / (72.0 - x/180 mL of water) = 14

14 = 180 * x / 180 * 72.0 - x

x = 14(72.0 - x)

x = 1008 - 14x

15x = 1008

x = 1008/15

x = 67.2 g

Weight of Y extracted from water in single extraction = 67.2 g

b) First extraction: Let x g of solute Y be extracted into methylene chloride

72.0 - x g of Y will be left in water

14 = x/90 / 72.0 - x/180

14 = 180x/ 90(72.0 - x)

14 = 2x/ 72.0 - x

2x = 14(72.0 - x)

2x =  1008 - 14x

16x = 1008

x = 63.0 g

Second extraction: Amount of solute left in water after first extraction = 72 - 63 = 9.0 g

Let x g of solute Y be extracted into methylene chloride

9.0 - x g of Y will be left in water

14 = x/90 / 9.0 - x/180

14 = 180x/ 90(9.0 - x)

14 = 2x/ 9.0 - x

2x = 14(9.0 - x)

2x =  126 - 14x

16x = 126

x = 7.90 g

Total weight of Y extracted = (63 + 7.90) g

Total weight of Y extracted from water in two successive extractions = 70.9 g

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer

answer seismograph;