Chemical Engineering - Mass Transfer - Discussion
Discussion Forum : Mass Transfer - Section 11 (Q.No. 5)
5.
1 kg of a saturated aqueous solution of a highly soluble component A at 60°C is cooled to 25°C. The solubility limits of A are (0.6 kgA/kg water) at 60°C and (0.2 kgA/kg water) at 25°C. The amount in kg of the crystals formed is
Discussion:
3 comments Page 1 of 1.
Zeeshan Ulhaq said:
5 years ago
0.6Kg of A/1.6kg of solution = 0.375 At 60° C.
0.2 kg of A/1.2 kg of solution = 0.166 At 25° C.
Crystal formed = 0.375 - 0.1666 = 0.2084.
0.2 kg of A/1.2 kg of solution = 0.166 At 25° C.
Crystal formed = 0.375 - 0.1666 = 0.2084.
(3)
Max said:
12 months ago
I think B should be the right answer.
OMB: F = L +C.
A bal: F*(0.6/1.6) = L (0.2/1.2) + C.
C = 0.25.
OMB: F = L +C.
A bal: F*(0.6/1.6) = L (0.2/1.2) + C.
C = 0.25.
(1)
Gaurav meena said:
5 years ago
I think the answer should be B.
(5)
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