Saturated Solution Concentrations and Osmotic Pressure Calculations

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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!
 
on Phys.org
You use too many significant digits every time.

First looks OK.

Second and third - approach seems correct (even if in the second your formula misses parantheses), but second opinion won't hurt.
 
Thank you Borek, and I agree, I'd still appreciate at least another person's opinion on my working.
 

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