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Elis [28]
2 years ago
5

The density of NaCl( s) is 2.165 g cm 3 at 25 C. How will the solubility of NaCl in water be affected by an increase in pressure

Chemistry
1 answer:
sammy [17]2 years ago
7 0

The solubility of NaCl in water will not be affected by an increase in pressure.

We know that the density of NaCl(s) in 2.165 g/cm³ at 25 °C and we want to know how will its solubility in water be affected when the pressure is increased.

<h3>What is solubility?</h3>

Solubility is the maximum mass of a solute that can be dissolved in 100 grams of solvent at a determined temperature.

The solubility of a solid, such as NaCl, in a liquid, is mainly affected by the temperature. However, since solids are not compressible, an increase in pressure will not affect its solubility.

On the other hand, the solubility of gases in water will increase with an increase in pressure, as stated by Henry's law.

The solubility of NaCl in water will not be affected by an increase in pressure.

Learn more about solubility here: brainly.com/question/11963573

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Five million gallons per day (MGD) of wastewater, with a concentration of 10.0 mg/L of a conservative pollutant, is released int
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Answer:

a) The concentration in ppm (mg/L) is 5.3 downstream the release point.

b) Per day pass 137.6 pounds of pollutant.  

Explanation:

The first step is to convert Million Gallons per Day (MGD) to Liters per day (L/d). In that sense, it is possible to calculate with data given previously in the problem.  

Million Gallons per day 1 MGD = 3785411.8 litre/day = 3785411.8 L/d

F_1 = 5 MGD (\frac{3785411.8 L/d}{1MGD} ) = 18927059 L/d\\F_2 =10 MGD (\frac{3785411.8 L/d}{1MGD} )= 37854118 L/d

We have one flow of wastewater released into a stream.  

First flow is F1 =5 MGD with a concentration of C1 =10.0 mg/L.

Second flow is F2 =10 MGD with a concentration of C2 =3.0 mg/L.  

After both of them are mixed, the final concentration will be between 3.0 and 10.0 mg/L. To calculate the final concentration, we can calculate the mass of pollutant in total, adding first and Second flow pollutant, and dividing in total flow. Total flow is the sum of first and second flow. It is shown in the following expression:  

C_f = \frac{F1*C1 +F2*C2}{F1 +F2}

Replacing every value in L/d and mg/L

C_f = \frac{18927059 L/d*10.0 mg/L +37854118 L/d*10.0 mg/L}{18927059 L/d +37854118 L/d}\\C_f = \frac{302832944 mg/d}{56781177 L/d} \\C_f = 5.3 mg/L

a) So, the concentration just downstream of the release point will be 5.3 mg/L it means 5.3 ppm.

Finally, we have to calculate the pounds of substance per day (Mp).  

We have the total flow F3 = F1 + F2 and the final concentration C_f. It is required to calculate per day, let's take a time of t = 1 day.  

F3 = F2 +F1 = 56781177 L/d \\M_p = F3 * t * C_f\\M_p = 56781177 \frac{L}{d} * 1 d * 5.3 \frac{mg}{L}\\M_p = 302832944 mg

After that, mg are converted to pounds.  

M_p = 302832944 mg (\frac{1g}{1000 mg} ) (\frac{1Kg}{1000 g} ) (\frac{2.2 lb}{1 Kg} )\\M_p = 137.6 lb

b) A total of 137.6 pounds pass a given spot downstream per day.

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