Extraction, Leaching and Adsorption - Mass Transfer - Chemical Engineering Previous Year Questions

It is desired to reduce the concentration of pyridine in 500 kg of aqueous solution from 20 weight percent to 5 wt percent in a single batch extraction using chloro-benzene as solvent. Equilibrium compositions (end points of the tie line) in terms of weight percent of pyridine-water-chlorobenzene are (5, 95, 0) and (11, 0, 89).
The amount of pure solvent required in kg for the operation is

A feed (F) containing a solute is contacted with a solvent (S) in an ideal stage as shown in the diagram below. Only the solute transfers into the solvent. The flow rates of all the streams are shown on a solute free basis and indicated by the subscript S. The compositions of the streams are expressed on a mole ratio basis. The extract leaving the contactor is divided into two equal parts, one part collected as the product (P) and the other stream is recycled to join the solvent. The equilibrium relationship is
\[ Y^* = 2X \]
The product flow rate (\( P_S \)) and composition (\( Y_{out} \)) are

A counter-current extraction column is designed to remove 99% of solute C from a solution of solvent A and solute C using pure solvent B. The initial concentration of solute in the solution of A + C is 20 wt %, and the total flow of solution is 1000 kg/h. If the equilibrium relationship is \(Y = 2X\), where \(Y\) = mass of C/mass of A and \(X\) = mass of C/mass of B. The minimum flow rate of solvent B required (in kg/h) is

Which of the following conditions are valid at the plait point?
P) Density difference between the extract and raffinate phases is zero
Q) Interfacial tension between the extract and raffinate phases is zero
R) Composition difference between the extract and raffinate phases is zero

In a batch adsorption process, 5 g of fresh adsorbent is used to treat 1 liter of an aqueous phenol solution. The initial phenol concentration is 100 mg/liter. The equilibrium relation is given by
\[ q^* = 1.3C \]
where \( q^* \) is the amount of phenol adsorbed in mg of phenol per gram of adsorbent; and \( C \) is the concentration of phenol in mg/liter in the aqueous solution.
When equilibrium is attained between the adsorbent and the solution, the concentration of phenol in the solution, rounded to 1 decimal place, is ______ mg/liter.

It is decided to extract A from a feed containing 20 mol% A and 80 mol% B in two ideal cross-current stages as shown below, using equal amount of pure solvent C in each stage. Components B and C are immiscible. 60% of A in the feed is extracted in Stage 1. The equilibrium relation is given by \( Y^* = 1.5 X \) where, \( X \) = moles of A per mole of B in raffinate, \( Y^* \) = moles of A per mole of C in extract in equilibrium with raffinate. The mol % of A in raffinate from Stage 2 is __________ (rounded off to second decimal place).

Two ideal cross-current stages operate to extract \(\mathbf{P}\) from a feed containing \(\mathbf{P}\) and \(\mathbf{Q}\), as shown below.

The mass flow rates of \(\mathbf{P}\) and \(\mathbf{Q}\) fed to Stage 1 are 1,000 kg h^-1 and 10,000 kg h^-1, respectively. Pure solvent (S) is injected at mass flow rates of 5,000 kg h^-1 and 15,000 kg h^-1 to Stages 1 and 2, respectively. The components \(\mathbf{Q}\) and \(\mathbf{S}\) are immiscible. The equilibrium relation is given by \(Y^* = 1.5 X\), where \(X\) is the mass of \(\mathbf{P}\) per unit mass of \(\mathbf{Q}\) in the raffinate, and \(Y^*\) is the mass of \(\mathbf{P}\) per unit mass of \(\mathbf{S}\) in the extract, which is in equilibrium with the raffinate. The mass flow rate of \(\mathbf{P}\) (in kg h^-1) in the raffinate from Stage 2 is __________ (round off to nearest integer).

Feed solution F is contacted with solvent B in an extraction process. Carrier liquid in the feed is A and the solute is C. The ternary diagram depicting a single ideal stage extraction is given below. The dashed lines represent the tie-lines.

The CORRECT option(s) is/are

500 mg of a dry adsorbent is added to a beaker containing 100 mL solution of concentration 100 mg phenol/L solution). The adsorbent is separated out after 5 h of rigorous mixing. If the residual concentration in the solution after separating the adsorbent is 30 mg phenol/L solution), the amount of phenol adsorbed (in mg per gram of dry adsorbent) is