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kateman

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**Q1:**A theoretical salt (X

_{2Y3) (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 ------------------------------------------------------------------- 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. ----------------------------------------------------------------- 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! }