Assume seawater is a solution of 0.570 M NaCl. Calculate the minimum pressure required at 10.00°C to desalinate seawater so that it contains no more than 177 mg NaCl per liter.

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princessesther2011 princessesther2011 · Answer 1 · 2020-02-26T03:05:10+01:00

Answer:

The minimum pressure required is 13.2 atm.

Explanation:

P = ∆CRT

Initial concentration of NaCl in sea water = 0.570 M

Final concentration of NaCl in sea water = concentration in g/L ÷ MW = 177 mg/L × 1 g/1000 mg ÷ (58.5 g/mol) = 0.00303 mol/L = 0.00303 M

∆C (change in concentration of NaCl in sea water) = 0.57 - 0.00303 = 0.56697 M

R is gas constant = 0.082057 L.atm/mol.K

T is temperature = 10 °C = 10+273 = 283 K

P (pressure) = 0.56697×0.082057×283 = 13.2 atm

andromache andromache · Answer 2 · 2022-02-15T16:54:02+01:00

The minimum pressure required is 13.2 atm.

We know that

P = ∆CRT

Also,

Initial concentration of NaCl in seawater = 0.570 M

Now

Final concentration of NaCl in sea water should be

= concentration in g/L ÷ MW

= 177 mg/L × 1 g/1000 mg ÷ (58.5 g/mol)

= 0.00303 mol/L = 0.00303 M

Now

∆C (change in concentration of NaCl in sea water)

= 0.57 - 0.00303

= 0.56697 M

Now

R is gas constant = 0.082057 L.atm/mol.K

And,

T is temperature = 10 °C = 10+273 = 283 K

So,

P (pressure) = 0.56697×0.082057×283

= 13.2 atm

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