can someone help me with the question number 2 is theanswer correct mass of laur

Question

can someone help me with the question number 2 is theanswer correct mass of lauric acid is 8.0016 g and mass of benzoic acidis 1.000 g , freezing temperature of pure lauric acid is 43.66degree c AND freezing point of acid lauric and mixture is 36.39degree c Calculate molality in mol/ kg using theformula T= Kf the KF VALUE for lauric acid is 3.9i did that and i got molality and t/kf = 1.9 mole kg second question it said and  im confusedhow we do it 2), find the mole of benzoic acid solute , usingthe molality and the mass in kg of lauric acid solvent please resolve thank you can someone help me with the question number 2 is theanswer correct mass of lauric acid is 8.0016 g and mass of benzoic acidis 1.000 g , freezing temperature of pure lauric acid is 43.66degree c AND freezing point of acid lauric and mixture is 36.39degree c Calculate molality in mol/ kg using theformula T= Kf the KF VALUE for lauric acid is 3.9i did that and i got molality and t/kf = 1.9 mole kg second question it said and  im confusedhow we do it 2), find the mole of benzoic acid solute , usingthe molality and the mass in kg of lauric acid solvent please resolve thank you

Explanation / Answer

Given mass of lauric acid is ,W = 8.0016 g
mass of benzoic acid is , w = 1.000 g molal depression constant = 3.9 freezing temperature of pure lauric acid is ,Ts = 43.66 degree c freezing point of acid lauric and mixture is , Tsol = 36.39 degree c We know T s -  T sol =T f              43.66 - 36.39 = T f          Tf = 7.27 degree c              43.66 - 36.39 = T f          Tf = 7.27 degree c The relation between Depression in freezing point &molality us given by T f = Kf * m                                                                                                               m = T f  / K f                                                                                                                    = 7.27 / 3.9                                                                                                                   = 1.8641 mol / Kg We know molality = no .of moles of solute / Kg of thesolvent So,  no .of moles of solute =molality * Kg of the solvent                                          = 1.8641 * 0.0080016                                          = 0.0149moles                                                                                                                    = 7.27 / 3.9                                                                                                                   = 1.8641 mol / Kg We know molality = no .of moles of solute / Kg of thesolvent So,  no .of moles of solute =molality * Kg of the solvent                                          = 1.8641 * 0.0080016                                          = 0.0149moles

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