Question

The following data was collected after running a calorimetric experiment using three salts.

The data collected for each experiment is as follows.

Experiment 1: 2.53 g of unknown salt added to 160. mL of water resulted in a ΔT value of −0.1°C.

Experiment 2: 2.53 g of unknown salt added to 160. mL of water resulted in a ΔT value of 0.7°C.

Experiment 3: 2.53 g of unknown salt added to 160. mL of water resulted in a ΔT value of −1.8°C.

Calculate the value of q for each of the experiments. (Assume that the heat capacity of water is 4.1801 J/g·°C.)

Based on the magnitude and sign of the q values calculated, determine the identity of the salt used in each experiment and the resulting ΔH_sol based on that deduction. (NOTE: You can also calculate ΔH_sol to see if your deduction makes sense when compared to the known values given above.)

The following data was collected after running a calorimetric experiment using three salts. Table 1: AH sol Values of Salts Salt AH gol (kJ/mol) | LiI -63.60 LINO3 -2.51 LiF 4.73 The data collected for each experiment is as follows. Experiment 1: 2.53 g of unknown salt added to 160. mL of water resulted in a AT value of -0.1°C. Experiment 2: 2.53 g of unknown salt added to 160. mL of water resulted in a AT value of 0.7°C. Experiment 3: 2.53 g of unknown salt added to 160. mL of water resulted in a AT value of -1.8°C. Calculate the value of q for each of the experiments. (Assume that the heat capacity of water is 4.1801 1/g.°C.) Experiment 1 497 Experiment 2 Experiment 3 497 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 sol based on that deduction. (NOTE: You can also calculate AM sol to see if your deduction makes sense when compared to the known values given above.) Experiment Identity of Salt AH sol (k]/mol) LII LINO3 10 LIF LII LINO LIF LII OLINO3 LIF

Answer 1

Expt1.

Heat change of solution = mass of solution × specific heat × temperature change

= (160+2.53) × 4.1801 × (-0.1)

= 162.53×4.1801 ×(-0.1)

= - 67.94 J.

Heat change in dissolution (q) = +67.94 J

Expt2.

Heat change in solution = 162.53×4.1801× (0.7)

= 475.57 J.

Heat change dissolution (q) = - 475.57 J.

Expt 3.

Heat change of solution

= 162.53×4.1801×(-1.8)

= 1223 J

Heat change of dissolution (q)

= - 1223 J.

Now moles of LiI (n) = (mass/molar mass) = (2.53/134) = 0.0189

Moles of LiF (n) = (mass/molar mass) = (2.53/26) = 0.097

Moles of LiNO₃ (n) = (2.53/69) = 0.0367

From the given values of $\Delta H_{sol}$ only LiF is positive

Now, for heat change of expt 2 has heat of dussolution positive , hence Expt 2 is LiF.

Now experimental value of heat of dissolution

$\Delta H_{sol}$ = (475.57/0.097) = 4902.7 J = 4.902 KJ/mole.

Now, For Expt 1.

If it is LiI

$\Delta H_{sol}$ = (q/n) = - (67.94/0.0189) = - 3594 J = -3.594 KJ.

If it is LiNO₃

$\Delta H_{sol}$ = - (67.94/0.0357) = - 1903 J = -1.903 KJ/mole

Theoretical value of $\Delta H_{sol}$ of LiNO₃ = - 2.51 KJ/mol

Hence, Expt 1 is LiNO​​​​​​₃.

For, expt 3

It is LiI

$\Delta H_{sol}$ =-(1223/0.0189) = - 64708 J = -64.71 KJ/mol

Now magnitude and sign of calculated value are

Experiment	identity	$\Delta H_{sol}$ (KJ/mol) (in 4 significant figure)
1	LiNO3	- 1.903
2	LiF	4.902
3	LiI	- 64.71