Chem

The date above s was obtained exploitation the following chassis . However, before apply 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 prepare that Therefore, has After that using the formula , it was found that , after that the unbelief was found using the above formula, therefore , therefore the entertain for s is .The it was the same principle for the rest of the results on tables. make band 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. 003632137 1212.037205.951.78339122297.650.0! 03359651 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 respectable essay, order it on our website: OrderEssay.net

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