Question

A "coffee-cup" calorimetry experiment is run for the dissolution of 2.5 g of lithium nitrate placed into 97.2 mL of water. The temperature of the solution is initially at 23.5 "C. After the reaction takes place, the temperature of the solution is 28.3 "C. Assume no heat is lost to the calorimeter.
1. Using a density of 1.0 g/ml for the waber added and adding in the mass of the iium nitrate, what is the total mass of the solution?
a. 99.7 g
b. 100 g
c. 94.7 g
d. 972 g
2. How much heat wasat sorted or lost byt, sumour digs? Use 4184 for the specific heat of the solution. Answer in units of kJ and make suro the sign conect.
a. 1.95 kJ
b. 2.00kJ
c. -1.90kJ
d. -201 kJ
3. How many moles of lithium nitrate are used in this experiment
a. 0.068 moles
b. 0.072 moles
c. 0.036 moles
d. 0.12 moles
4. What would be the enthalpy for the dissolution reaction of one mole of ithum nitrate? Answer in killimoll and watch the sign tor the enthalpy
a. 27.1 kJ/mole
b.556 kJ/mole
c. -27.1 kJ/mole
d. -55.6 kJ/mole
5. If the lid was left off during the temperature measnments nts e permet row would affect the value of a?
a. q would be higher
b. would be unaffected
c. qwould be low

Answer 1

Answer:

1) 99.7 g is the total mass of the solution.

2)The amount of heat absorbed by solution is 2.00 kJ.

3) Moles of lithium nitrate are used is 0.036 mol.

4) The enthalpy for the dissolution reaction of one mole of lithium nitrate is -55.6 kJ/mol.

5) The value of q would be low.

Explanation:

1) mass of the water = M

Volume of the water,V =  97.2 mL

Density of the water ,d= 1.0 g/mL

$Mass=Volume \times Density$

$M=d\times V=1.0 g/ml\times 97.2 ml=97.2 g$

Mass of the lithium nitrate = M' = 2.5 g

Mass of the solution ,m = M + M' = 97.2 g + 2.5 g = 99.7 g

2) Mass of solution = m = 99.7 g

Specific heat of the solution = c = 4.184 J/g°C

Change in temperature = ΔT = 28.3°C - 23.5°C = 4.8°C

Temperature of the solution is increased which means that heat absorbed by the solution.

Heat energy absorbed by the solution = Q

$Q=mc\Delta T$

$=99.7 g\times 4.184 J/g^oC\times 4.8^oC=2,002.30 J=2.002 kJ\approx 2.00 kJ$

The amount of heat absorbed by solution is 2.00 kJ.

3) Mass of lithium nitrate = 2.5 grams

Molar mass of lithium nitrate = 69 g/mol

Moles of lithium nitrate , n= $\frac{2.5 g}{69 g/mol}=0.036 mol$

Moles of lithium nitrate are used is 0.036 mol.

4) Moles of lithium nitrate = n = 0.036 mol

Heat absorbed by solution = heat released on dissolution = 2.00 kJ.

heat released on dissolution = -2.00 kJ

(negative sign indicates that heat is released)

Enthalpy of the dissolution reaction of one mole of lithum nitrate:

$\Delta H_{dissolution}=\frac{-Q}{n}=\frac{2.00 kJ}{0.036 mol}=-55.6 kJ/mol$

The enthalpy for the dissolution reaction of one mole of lithium nitrate is -55.6 kJ/mol.

5) If the lid was left off during the temperature measurement the the heat absorbed by the solution will lost some amount of heat to the atmosphere which will increase the temperature the solution and will also effect the other observations.