Chem

The date above s was obtained exploitation the following prototype . However, before use this formula we have to convert grams of to moles using the stoichiometry , it was effectuate that Moles of = moles of Then using the formula for question , it was plant that Therefore, has After that using the formula , it was found that , after that the suspense was found using the above formula, therefore , therefore the shelter for s is .The it was the same principle for the rest of the results on tables. frivol down of (g) Mass of upshot (g)S(mol/kg)Ln sT(K)1/T(K) 99.095 20.28 4.438 0.003 1.4902 0.003 299.95 0.1 0.00334 1.11 1212.095 20.28 5.917 0.004 1.7778 0.004 309.45 0.1 0.003232 1.04 1515.095 20.28 7.347 0.004 1.9942 0.004 316.55 0.1 0.003159 0.998 1818.095 20.28 8.826 0.004 2.1777 0.004 322.45 0.1 0.003101 0.062 1M KClMass of (g) Mass of solvent (g)S(mol/kg)Ln sT(K)1/T(K) 99.006 19.7 4 .52 5.4* 1.51 5.4* 309.65 0.1 0.0033 1*10^-6 1212.006 19.7 6.04 7.3* 1.8 7.3* 316.35 0.1 0.0032 1*10^-6 1515.006 19.7 7.51 9.1* 2.02 9.1* 323.85 0.1 0.0031 0.9*10^-6 1818.006 19.7 9.04 10.9* 2.2 10.9* 329.85 0.1 0.003 0.

9*10^-6 2M KClMass of (g) Mass of solvent (g)S(mol/kg)Ln sT(K)1/T(K) 99.03620.0874.4511.493313.250.00319 1212.04920.0875.931.78322.350.003102 1515.05720.0877.4132.003329.850.003022 1818.05820.0878.892.185336.250.002974 uncertainty0.0010.0015.4,5.74,6.15,6.62 (*10^-4)5.4,5.74,6.15,6.62 (*10^-4)0.11.02.962,.919,.884 (*10^-6) 1M NaClMass of (g) Mass of solvent (g)S(mol/kg)Ln sT(K)1/T(K) 99.028204.461.495148766275.320.00 3632137 1212.037205.951.78339122297.650.003! 359651 1515.048207.442.006870849300.450.003328341 1818.054208.932.189416395308.050.003246226 uncertainty0.0010.1.05,.07,.06,.075.05,.07,.06,.0750.11.32,1.13,1.1,1.06 (*10^-6) 2M NaClMass of (g)...If you emergency to get a wide-cut essay, order it on our website: OrderEssay.net

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