Question

The following data was collected after running a calorimetric experiment using three salts. Table 1: AHsol Values of Salts Salt AHsol (kJ/mol) LiI -63.60 LINO3 -2.51 LiF 4.73 The data collected for each experiment is as follows. Experiment 1: 2.66 q of unknown salt added to 180. mL of water resulted in a AT value of -0.1°c. Experiment 2: 2.66 q of unknown salt added to 180. mL of water resulted in a AT value of 0.6°C Experiment 3: 2.66 q of unknown salt added to 180. mL of water resulted in a AT value of -1.7°C. Calculate the value of q for each of the experiments. (Assume that the heat capacity of water is 4.1801 J/g.oC.) 4.0-75.2418 Experiment 1 4.0 Experiment 2 J 4.0 Experiment 3 J Based on the magnitude and sign of the q values calculated, determine the identity of the salt used in each experiment and the resulting AH based on that deduction. (NOTE: You can also calculate AHsol to see if your deduction makes sense when compared to the known values given above.) Experiment Identity of Salt AHsol (kJ/mol) OLII 1 O LINO3 4.0 LiF 2 LiI O LINO3 4.0 OLIF 3 OLi O LINO3 4.0 LiF ООс

Answer 1

(Sol) Criven in the question Mass of salt = 2.669 Volume of water - 180 ml Mass of water = Volume x density - 180X = 180g Total mass - 180+ 2.66 = 182.66 g (p= 4.1801 Ilic Now Experiment I AT = -0.1°C q = mx CP XOT = 182.66 X 4.180 1X (-0112 - 76.35 I Experiment -2 ATT 0.6°C q= mx cpXOT = 182.66 X 4.1801 x (0.6) = 458.12 Experiment of Ot= a1.9°C q = mx CP XOT = 182.66 X 4.1801 x(-1,7) 7- 1298.0 In Molar mass of LiI = 134 glmal Molar man of LiNO3 = 69 g / mal Molar mass of lif = 268 l mal No. of moles of lit = 9.6.6/134 = 19, 8 X 10 moles. No. of moles of LiNO₃ = 8.66/69 = 38.5 x 10 moles No. of moles of LiF= 2.66/26 = 109.3x10 3 moles According to given Seeble only lif will have positive valus. -

So if lif s present on= q in experiment -2: Noof moles of Lif = 458 12 102.3x10-3 = 4.478 X 103 Ilmal in Dne = 4.478 kJ / mal s Now comparing other two Lit will have higher value of an So for lil on = q in expr enos No. of moles of LI I 5 -76.35 /19.8 x16 FN 53-856 x I/mol " for LiNO. = a in, exp-3 an t No. of maly of lino3 b > -1298.01 / 38.5X10 Now Identity of salt o = -33.714 KJ/mol Exp-1 Litr on - - 3 85.6 KIlmol Exp-2 LIF ,OH = 4.478 kJ/mol La 18 Exp3 → LINO3 non a-33.714 Il mal