In a liquid-liquid extraction, 10 kg of a solution containing 2 kg of solute C and 8 kg of solvent A is brought into contact with 10 kg of solvent B. Solvent A and B are completely immiscible in each other whereas solute C is soluble in both the solvents. The extraction process attains equilibrium. The equilibrium relationship between the two phases is Y* = 0.9X, where Y* is the kg of C/kg of B and X is kg of C/kg of A. Choose the correct answer.

We are given:

Initial solution: 10 kg containing

Solute C: 2 kg

Solvent A: 8 kg

Contacted with: 10 kg of solvent B

A and B are immiscible

Solute C is distributable between A and B

Equilibrium relationship:

𝑌
∗
=
0.9
𝑋
Y
∗
=0.9X
where:
𝑋
=
kg C
kg A
X=
kg A
kg C
​

𝑌
∗
=
kg C
kg B
Y
∗
=
kg B
kg C
​


Let:

𝐶
𝐴
C
A
​
: kg of C remaining in solvent A

𝐶
𝐵
C
B
​
: kg of C extracted to solvent B

Total C =
𝐶
𝐴
+
𝐶
𝐵
=
2
 
kg
C
A
​
+C
B
​
=2kg

Also:

𝑋
=
𝐶
𝐴
8
X=
8
C
A
​

​


𝑌
∗
=
𝐶
𝐵
10
Y
∗
=
10
C
B
​

​


From the equilibrium relationship:

𝐶
𝐵
10
=
0.9
⋅
𝐶
𝐴
8
10
C
B
​

​
=0.9⋅
8
C
A
​

​

𝐶
𝐵
=
0.9
⋅
10
8
⋅
𝐶
𝐴
=
1.125
⋅
𝐶
𝐴
C
B
​
=0.9⋅
8
10
​
⋅C
A
​
=1.125⋅C
A
​

Substitute into mass balance:

𝐶
𝐴
+
𝐶
𝐵
=
2
⇒
𝐶
𝐴
+
1.125
⋅
𝐶
𝐴
=
2
⇒
2.125
⋅
𝐶
𝐴
=
2
⇒
𝐶
𝐴
=
2
2.125
≈
0.9412
 
kg
C
A
​
+C
B
​
=2⇒C
A
​
+1.125⋅C
A
​
=2⇒2.125⋅C
A
​
=2⇒C
A
​
=
2.125
2
​
≈0.9412kg
Then:

𝐶
𝐵
=
2
−
0.9412
=
1.0588
 
kg
C
B
​
=2−0.9412=1.0588kg
✅ Final Answer:
B. Less than 2 kg but more than 1 kg of C is transferred to solvent B

We are given:

Initial solution: 10 kg containing

Solute C: 2 kg

Solvent A: 8 kg

Contacted with: 10 kg of solvent B

A and B are immiscible

Solute C is distributable between A and B

Equilibrium relationship:

𝑌
∗
=
0.9
𝑋
Y
∗
=0.9X
where:
𝑋
=
kg C
kg A
X=
kg A
kg C
​

𝑌
∗
=
kg C
kg B
Y
∗
=
kg B
kg C
​


Let:

𝐶
𝐴
C
A
​
: kg of C remaining in solvent A

𝐶
𝐵
C
B
​
: kg of C extracted to solvent B

Total C =
𝐶
𝐴
+
𝐶
𝐵
=
2
 
kg
C
A
​
+C
B
​
=2kg

Also:

𝑋
=
𝐶
𝐴
8
X=
8
C
A
​

​


𝑌
∗
=
𝐶
𝐵
10
Y
∗
=
10
C
B
​

​


From the equilibrium relationship:

𝐶
𝐵
10
=
0.9
⋅
𝐶
𝐴
8
10
C
B
​

​
=0.9⋅
8
C
A
​

​

𝐶
𝐵
=
0.9
⋅
10
8
⋅
𝐶
𝐴
=
1.125
⋅
𝐶
𝐴
C
B
​
=0.9⋅
8
10
​
⋅C
A
​
=1.125⋅C
A
​

Substitute into mass balance:

𝐶
𝐴
+
𝐶
𝐵
=
2
⇒
𝐶
𝐴
+
1.125
⋅
𝐶
𝐴
=
2
⇒
2.125
⋅
𝐶
𝐴
=
2
⇒
𝐶
𝐴
=
2
2.125
≈
0.9412
 
kg
C
A
​
+C
B
​
=2⇒C
A
​
+1.125⋅C
A
​
=2⇒2.125⋅C
A
​
=2⇒C
A
​
=
2.125
2
​
≈0.9412kg
Then:

𝐶
𝐵
=
2
−
0.9412
=
1.0588
 
kg
C
B
​
=2−0.9412=1.0588kg
✅ Final Answer:
B. Less than 2 kg but more than 1 kg of C is transferred to solvent B

Less than 2 kg but more than 1 kg of C is transferred to solvent B

could you show the solution for this

How di you find the answer can you show you solution

Answer is 1.0588 kg of solute is transferred.