- #1
kateman
- 114
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There are 3 questions below with my working and answers, could someone please check that I am right, even if you can only give approval or correction on just one question and nothing else - anything would be appreciated. Please forgive my not following the template, as I didn't think it would work with checking correct answers. Thanks for any responces!
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Q1: A theoretical salt (X2Y3) (with Ksp= 6.26 x 10 -15 ) is in a saturated solution. What is its millimolar concentration (mM) of X?
A1: Ksp= [X]2 [Y]3 = 6.26 x 10 -15
therefore concentration of Y = 3/2 concentration of X (or is it 2/3, iam not sure?)
therefore rewrite Y as [3/2 X]3
therefore Ksp= [X]2x 27/8 X3
= 27/8 [X]5
therefore, with rearranging, [M] = 1.131514468x10-3 mol/L
= 1.131514468 mM/L
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Q2: A reaction at temperature of 287K gives a change in enthaply of -7KJ and a free energy change of -10kJ, what is the change in entropy for this reaction (in J/K)?
A2: [tex]\Delta[/tex]G = [tex]\Delta[/tex]H - T[tex]\Delta[/tex]S
with rearranging: [tex]\Delta[/tex]S = [tex]\Delta[/tex]H - [tex]\Delta[/tex]G / T
=(-7 - -10)/287 = 0.010452961 J/K
That doesn't seem right to me.
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Q3: If a 0.5 L solution at a temperature of 294.75 K contains 13.7g of an unknown solute (with a molecular mass of 60.094g), what is its osmotic pressure in atmospheres?
A3: since moles (n) = concentration (c)/ volume (v) = mass (m)/ molecular weight (M)
then C = m/Mv = 13.7 / (60.094x0.5) = 0.455952341 mol/L
Now use the values of C, T and the gas constant (8.314 J/mol K) into the osmotic pressure formula P=CRT
P= 0.455952341x8.314x294.75 = 1117.334694 Pa (am I right to say that its in pascals, or is it in killapascals [Kpa]?)
atmospheres = 101.325 Kpa = 101325 Pa
osmotic pressure (in atmospheres) = 1117.334694/101325 = 0.011027236 atms
This is something iam really not sure about, the answer seems almost wrong to me but iam not sure where I would have gone wrong in my working, can someone please help? It would be very much appreciated!
Thank you!
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Q1: A theoretical salt (X2Y3) (with Ksp= 6.26 x 10 -15 ) is in a saturated solution. What is its millimolar concentration (mM) of X?
A1: Ksp= [X]2 [Y]3 = 6.26 x 10 -15
therefore concentration of Y = 3/2 concentration of X (or is it 2/3, iam not sure?)
therefore rewrite Y as [3/2 X]3
therefore Ksp= [X]2x 27/8 X3
= 27/8 [X]5
therefore, with rearranging, [M] = 1.131514468x10-3 mol/L
= 1.131514468 mM/L
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Q2: A reaction at temperature of 287K gives a change in enthaply of -7KJ and a free energy change of -10kJ, what is the change in entropy for this reaction (in J/K)?
A2: [tex]\Delta[/tex]G = [tex]\Delta[/tex]H - T[tex]\Delta[/tex]S
with rearranging: [tex]\Delta[/tex]S = [tex]\Delta[/tex]H - [tex]\Delta[/tex]G / T
=(-7 - -10)/287 = 0.010452961 J/K
That doesn't seem right to me.
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Q3: If a 0.5 L solution at a temperature of 294.75 K contains 13.7g of an unknown solute (with a molecular mass of 60.094g), what is its osmotic pressure in atmospheres?
A3: since moles (n) = concentration (c)/ volume (v) = mass (m)/ molecular weight (M)
then C = m/Mv = 13.7 / (60.094x0.5) = 0.455952341 mol/L
Now use the values of C, T and the gas constant (8.314 J/mol K) into the osmotic pressure formula P=CRT
P= 0.455952341x8.314x294.75 = 1117.334694 Pa (am I right to say that its in pascals, or is it in killapascals [Kpa]?)
atmospheres = 101.325 Kpa = 101325 Pa
osmotic pressure (in atmospheres) = 1117.334694/101325 = 0.011027236 atms
This is something iam really not sure about, the answer seems almost wrong to me but iam not sure where I would have gone wrong in my working, can someone please help? It would be very much appreciated!
Thank you!