Question

A 72.0 g sample of an organic solid is dissolved in 180mL of water. The solid is extracted using one 60 mL extraction in the first extraction of an organic solvent which has a partition (distribution) coefficient with water of 10. The first extraction removed 55.4 g of solid from water. What are the numbers that need to go in box A and B to calculate the volume of solvent (y) that would be necessary to remove an additional 7.0g from the remaining sample dissolved in water. You DON’T have to complete the calculation to solve for y.

Answers

  1. Answer:

    V_{7.0}\approx 235ml

    Explanation:

    From the question we are told that

    mass of sample M=72.0 grams

    volume of water V=180 mL

    volume for extraction V'=60mL

    partition (distribution) coefficient water d=10

    initial extraction removal x=55.4g

    Generally the equation for the weight of sample x_n  is mathematically given by

    x_n=x*(\frac{DV}{DV+V'})^n

    x_n=55.4(\frac{10*180}{10*180+60})^1

    x_n=53.613g

    Generally the weight extracted x_e is therefore

    w_e=x-x_n

    w_e=55.4-53613

    w_e=1.787

    w_e=1.787 is extracted with 60ml solvent .

    Therefore volume of solvent (y) that would be necessary to remove an additional 7.0g

    V_{7.0}=\frac{60}{1.767}*7

    V_{7.0}=235.030ml

    V_{7.0}\approx 235ml

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