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One hundred milliliters per minute of 5M NaOH solution (sp. gr. = 1.18) is mixed with 10M NaOH solution (sp. gr. = 1.37). It is desired to produce a solution containing 11.7 % mole NaOH. Calculate the required volumetric flow rate (ml/min) of 10M NaOH solution. Also, calculate the volumetric flow rate at the final solution. (Atomic Weights: Na = 23, O = 16, H = 1)
Solution:
The given word problem is about mixing of two NaOH solutions with different concentrations to produce a desired NaOH solution which involves the principles of Stoichiometry. The total amount of a substance in the reactants or incoming ingredients must be equal to the total amount of a substance in the final products. In short, the Law of Conservation of Mass must be followed all the time. To illustrate the given problem, it is better to draw the flow diagram as follows
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Basis: 100 ml/min of 5M NaOH Solution
Let x = be the volumetric flow rate of 10M NaOH in ml/min
Molecular Weight Data:
NaOH = 40
H2O = 18
Consider 5M NaOH Solution:
The total moles of NaOH from the two solutions is equal to
The total moles of H2O from the two solutions is equal to
Hence, the % mole of NaOH is equal to
Material Balance of NaOH:
The volumetric flow rate of Final Solution is 100 ml/min + 78.2 ml/min = 178.2 ml/min.