The answer
first of all, we should know that NaOH is a strong base. For such a product, the conentration of the OH - is equivalent to the concentration of the NaOH itself.
that means:
[ OH -] = [ NaOH] =<span>0.001 62
and for a strong basis, pH can be calculated as pH = 14 + log </span>[ OH -]
first we compute log [ OH -] :
log [ OH -] = log (0.001 62)= -2.79
finally pH = 14 -2.79 = 11.20
Step 1
The osmotic pressure is calculated as follows:

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Step 2
<em>Information provided:</em>
The mass of solute = 13.6 g
Volume of solution = 251 mL
Absolute temperature = T = 298 K
The molar mass of solute = M = 354.5 g/mol
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Step 3
Procedure:
1 L = 1000 mL => Volume = 251 mL x (1 L/1000 mL) = 0.251 L
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C = moles of solute/volume of solution (L)
C = mass of solute/(molar mass x Volume (L))
C = 13.6 g/(354.5 g/mol x 0.251 L)
C = 0.153 mol/L
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π = C x R x T
π = 0.153 mol/L x 0.082 atm L/mol K x 298 K
π = 3.74 atm
Answer: π = 3.74 atm
Answer:C The final product cannot be converted back to the original ingredients.