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jolli1 [7]
3 years ago
8

Consider two solutions separated by a semipermeable membrane, as shown in the illustration to the right. The membrane allows the

passage of small molecules and ions, but not large molecules like polysaccharides or proteins. Solution A contains a 10% solution composed
of glucose and the protein albumin dissolved in water.
Solution B contains a 5% solution of NaCl in water. Indicate whether each substance in the system would flow into Solution A, Solution B, or neither.

1. Water
2. NaCl
3. glucose
4. Albumin
5. Glucose

Chemistry
1 answer:
Evgen [1.6K]3 years ago
6 0

Answer:

The correct movement would be -

1. Water  - into solution A.

2. NaCl  - into solution A.

3. glucose  - into Solution B.

4. Albumin  - neither.

Explanation:

All the substances are separated by the semipermeable membrane and the semipermeable membrane allows the only small molecule to pass through it. So the movement of the given substance would be -

1. Water  - into solution A.

Water molecules are small and can easily pass through the semipermeable membrane as it is given that the solution b has low solute concentration and solution A has high solute concentration. It is known that the movement of the solvent always takes place from low solute concentration to high so the movement of water will be into solution A.

2. NaCl  - into solution A.

The movement of small ionic molecule NaCl is always from high to low concentration as it is given that solution B has high concentration than solution A so movement will take place into solution A.

3. glucose  - into Solution B.

It is also a small molecule and moves from the high glucose region to the low glucose concentration region, in solution A the concentration of glucose is high than solution B so movement would be into solution B.

4. Albumin  - neither.

Albumin is a protein which is macromolecule and large in size to pass through the semipermeable membrane so, albumin move neither solution A nor solution B.

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What is the entropy of this collection of training examples with respect to the positive class B. What are the information gains
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$P(-) = \frac{5}{9}$

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b). For the attribute all the associating increments and the probability are :

  $a_1$   +   -

  T   3    1

  F    1    4

Th entropy for   $a_1$  is given by :

$\frac{4}{9}[ -\frac{3}{4}\log\left(\frac{3}{4}\right)-\frac{1}{4}\log\left(\frac{1}{4}\right)]+\frac{5}{9}[ -\frac{1}{5}\log\left(\frac{1}{5}\right)-\frac{4}{5}\log\left(\frac{4}{5}\right)]$

= 0.7616

Therefore, the information gain for $a_1$  is

  0.9911 - 0.7616 = 0.2294

Similarly for the attribute $a_2$  the associating counts and the probabilities are :

  $a_2$  +   -

  T   2    3

  F   2    2

Th entropy for   $a_2$ is given by :

$\frac{5}{9}[ -\frac{2}{5}\log\left(\frac{2}{5}\right)-\frac{3}{5}\log\left(\frac{3}{5}\right)]+\frac{4}{9}[ -\frac{2}{4}\log\left(\frac{2}{4}\right)-\frac{2}{4}\log\left(\frac{2}{4}\right)]$

= 0.9839

Therefore, the information gain for $a_2$ is

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$a_3$     Class label      split point       entropy        Info gain

1.0         +                        2.0            0.8484        0.1427

3.0        -                         3.5            0.9885        0.0026

4.0        +                        4.5            0.9183         0.0728

5.0        -

5.0        -                        5.5            0.9839        0.0072

6.0        +                       6.5             0.9728       0.0183

7.0        +

7.0        -                        7.5             0.8889       0.1022

The best split for $a_3$  observed at split point which is equal to 2.

c). From the table mention in part (b) of the information gain, we can say that $a_1$ produces the best split.

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