If 2.15 g of an unknown solid nonelectrplyte is dissolved in18.1 g of naphthalen

Question

If 2.15 g of an unknown solid nonelectrplyte is dissolved in18.1 g of naphthalene, the resulting solution is foundexperimentally to freeze 4.3 C lower than pure naphthalene. If thefreezing point depression constant for naphthalene is 6.85 C/m,calculate each of the following. a. The molality of the solution by using only the freezingpoint depression and the freezing point depression constant. b. The number of moles of unknown dissolved in the 10.1g ofnaphthalene using the mass of naphthalene and the molality of theunknown. c. The molar mass of the unknown. d. The molecular formula of the unknown if it is comprised of40.0% carbon, 6.7% hydrogen and 53.3% oxygen. e. Why is it important to specify that the unkwnon is anonelectrolyte? Thank you!! If 2.15 g of an unknown solid nonelectrplyte is dissolved in18.1 g of naphthalene, the resulting solution is foundexperimentally to freeze 4.3 C lower than pure naphthalene. If thefreezing point depression constant for naphthalene is 6.85 C/m,calculate each of the following. a. The molality of the solution by using only the freezingpoint depression and the freezing point depression constant. b. The number of moles of unknown dissolved in the 10.1g ofnaphthalene using the mass of naphthalene and the molality of theunknown. c. The molar mass of the unknown. d. The molecular formula of the unknown if it is comprised of40.0% carbon, 6.7% hydrogen and 53.3% oxygen. e. Why is it important to specify that the unkwnon is anonelectrolyte? Thank you!!

Explanation / Answer

Given mass of unknown , m' = 2.15 g
mass of the solvent naphthalene , M = 18.1 g = 0.0181 Kg Depression in freezing point , Tf = 4.3 oC freezing point depression constant for naphthalene , Kf= 6.85 C/m We know that Tf = Kf * m Where m = molaity of the solution So , m = Tf / Kf            =0.6277 m Therefore molality of the solution = 0.6277 m But molality , m = ( mass / Molar mass of unknown ) / mass ofthe solvent in Kg                 0.6277 = ( 2.15 / m'' ) / 0.0181                        m'' = 189.2377 g So molar mass of unknown is 189.2377 g molality , m = no .of moles of solute, n / mass ofsolvent                  n = m* mass of solvent                    = 0.6277 m * 0.0101 Kg                    = 0.00634 moles So required moles = 0.00634 mol The molecular formula of the unknown if it is comprised of40.0% carbon, 6.7% hydrogen and 53.3% oxygen. No . of moles of C = 40 / 12 = 3.33 No . of moles of H = 6.7 / 1 = 6.7 No .of moles of O = 53.3 / 16 = 3.33 The least moles is 3.33.Divide all the three moles by 3.33 weget 3.33 / 3.33 , 6.7/3.3 , 3.33/3.33                                                                                                    1, 2,1 for C , H & O respectively So the emperical formula is CH2O Emperica;l formula mass of CH2O is 12 + 2 * 1 + 16 = 30g Molar mass of the unknown is 189.23 So Molar mass / Emperical formula   = 189.23 /30 = 6.3 ~ 6 So the Molecular formula of the compound is 6 ( CH2O )                                                                    = C6H12O6 So required moles = 0.00634 mol The molecular formula of the unknown if it is comprised of40.0% carbon, 6.7% hydrogen and 53.3% oxygen. No . of moles of C = 40 / 12 = 3.33 No . of moles of H = 6.7 / 1 = 6.7 No .of moles of O = 53.3 / 16 = 3.33 The least moles is 3.33.Divide all the three moles by 3.33 weget 3.33 / 3.33 , 6.7/3.3 , 3.33/3.33                                                                                                    1, 2,1 for C , H & O respectively So the emperical formula is CH2O Emperica;l formula mass of CH2O is 12 + 2 * 1 + 16 = 30g Molar mass of the unknown is 189.23 So Molar mass / Emperical formula   = 189.23 /30 = 6.3 ~ 6 So the Molecular formula of the compound is 6 ( CH2O )                                                                    = C6H12O6

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