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seraphim [82]
3 years ago
9

Evaporation of Cane Sugar Solutions. An evaporator is used to concentrate cane sugar solutions. A feed of 10 000 kg/d of a solut

ion containing 38 wt % sugar is evaporated, producing a 74 wt % solution. Calculate the weight of solution produced and amount of water removed.
Chemistry
1 answer:
Vera_Pavlovna [14]3 years ago
8 0

Answer:

Weight of solution produced = 5135 kg

Amount of water removed = 4865 kg

Explanation:

For the balance of mass, the incoming mass of sugar must be equal to the outgoing mass. So, the incoming mass (mi) is 38% of 10000 kg

mi = 0.38x10000 = 3800 kg

The outgoing mass (mo) must be 3800 kg, and it is 74% of the total mass (mt)

mo = 0.74xmt

0.74xmt = 3800

mt = 3800/0.74

mt = 5135 kg

This is the mass of solution produced.

The amount of water removed (wr) is the amount of water incoming (wi) less the amount of water outgoing (wo). Both will be the total mass less the mass of sugar :

wi = 10000 - 3800 = 6200 kg

wo = 5135 - 3800 = 1335 kg

wr = wi - wo

wr = 6200 - 1335

wr = 4865 kg

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Answer:

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c. The [H⁺] is 1.74*10⁻⁴ M.

d. The [OH⁻] is 1.51*10⁻⁵ M.

Explanation:

pH is a measure of acidity or alkalinity that indicates the amount of hydrogen [H⁺] or hydronium [H₃O⁺] ions present in a solution or substance. Mathematically the pH is the negative logarithm of the molar concentration of the hydrogen or proton ions (H⁺) or hydronium ions (H₃O⁺):

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You know [H⁺]= 3.8*10⁻⁴ M. Then:

pH= -log (3.8*10⁻⁴)

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<em>The pH of a sample of lemon juice with [H⁺] = 3.8*10⁻⁴ M is 3.4, indicating acidity.</em>

<u><em>Case b</em></u>

Like pH, pOH (or OH potential) is a measure of the basicity or alkalinity of a solution.

The pOH indicates the concentration of hydroxyl ions [OH-] present in a solution:

pOH= -log₁₀ [OH⁻]= -log [OH⁻]

The pOH is related to the pH by the formula:

pH + pOH= 14

In this case you know [OH⁻]=1.9*10⁻⁶ M. Then:

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pH= 8.28

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You know pH=3.76. Then:

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[H⁺]=10^{-3.76}

[H⁺]= 1.74*10⁻⁴ M

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You know that pH + pOH=14 and pH is 9.18. Then

9.18 + pOH=14

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pOH= 4.82

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4.82= -log [OH⁻]

-4.82= log [OH⁻]

[OH⁻]=10^{-4.82}

[OH⁻]= 1.51*10⁻⁵ M

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