Question

ed your results in Table 14.5 3 Use the following steps to determine the perce following steps to determine the percent of the treening point depressions you determined Sperimentally from the expected freezing point de u te and record your results in Table 14.5 Show sample calculations for your sodium chloride solution. Use the actual molar mass of each solute to determi edted molality of 0.234 g of each solute in 4.00 x 10-3 kg of water. Sample Calculation: mass-0.2345 mm Nacl = 59.44m Wea: 0.23490 .2344 0.00290M > 0.00390/4.00 Test Tubes 0.234. se 0. 0.004 moes > 0.004 58.440 .00 miles /4.001 00 mm 60.0603 60.00 0.00390m 1 0.00390m/4.00*10-3= 0.975m (4 Glucose -0.234901234g Dilsug 01234 -0.00130m 1400*10-3 -0.325 0 0120m 14003 mm=180.lleg 180ollega b Determine the expected freezing point depression for a solution with this molality Sample Calculation: Mac 21.56*1 = 3.72 Unknown A (urea) ATE=1.kfim NA 2 . * 1.800.975=1.81 unknown B (Glucose) 181.86* 0.325 = 0.605 14 c Determine the percent error of the freezing point depressions you determined experimentally from the expected freezing point depression for each solute. Sample Calculation: No expected - acceptable *100 Glucose -71.8 acceptable 0.605-2, 1453 *100=71.8 2.1453 Nach 3.72-2.23293*100=66.6% 2.23293 vrea. 81 - 2.62978 *100 = -31.2% 2.62978

#3 part B: the eqution tf=ikfm

how do i get rid of the kg so i just have celsius left over or am i doing this completely wrong?

Answer 1

given that mass of each solute = 0.2348 mass of water = 4.00x103 kg we know that molality = moles of mass of solute solvent in kg- unit of molality - and moles of solute = mass of solute .. Molar mass of solute For Solyte Nach Mass of Solute dach = 0.234 g Molar mass of solute dach = 58.44 almal moles of soluze daci = 0.234.8 58.44 med moles of solute dacl - 0.00400 mol Similarly, моѕѕ оf uъео — о. зЧ 2 Molay mass of area = 60.06 Ilmol i - Modes of solute -0.2348 area = - Modes of solute urea - 0.0039 Moll

arass modar of mass Glucose = 0.234 g of glucose = 180.16 g/mol месо съсе — 0 — б. 23 ч 280.16 mol molality of dach solution = moles of dalk mass of water (kg) 1m 1.o mol = kg oooh mol 4.00x103 kg dace solution of = 1.0 mol/kg = 1 m molality dow, molality of urea solution = mates of urea - mass of water (kg) - molality of sy moe - 10.975ml urea = solution 0.0039 mol - To 4.00 x 20 3 kg glucose moles of glucose mass of water (in kg) morality of - 0.00130 mol 4.00x 10-3ko = 0.325 man I molality of glucose = 0.325 mot hozy . 325 ml = {0.325 m

o For Solution - freezing can be point given depression as follows: - ДТ - СКем ا عمالي i van't hoFF Fach KE ~ Freeting point depression KF = 1.86 oclon = 1.86 oc/morkg! m = molality - mos = m constant For water now. For daca - ATE = 2x1.86 ocloh x 1m = 3.72 oC - (- = 2 (dat, at 7 - Hence, for dace, OTF = 3-7200 / for urea, i=1.0 - ATE KE M = 1x 1.86°clxo.975 ATE=11.81°C for urea : for glycose i ATE = 1X ( 86 alex0.325 of = 0.605°c - - ATE = 1 x 1.86 •clorot. Kgit x8.325 rol kigt ATE = 10.605°C for glucose