A well insulated calorimeter containing 100 ml of water was calibrated by dissolving a known amount of sodium chloride in the calorimeter. The dissolution of sodium chloride in water can be represented by the equation: NaCl(s) + aq Na*(aq) + Cl(aq) AH=+3.90 kJ mol-¹ The temperature of the calorimeter was initially 18.00 °C. A sample of 2.55 g of sodium chloride was dissolved in the 100 mL and the temperature was recorded as 16.30 °C. The calibration factor of the calorimeter is (express your answer to 2 decimal places) 0.42 kJ °C-1

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A well insulated calorimeter containing 100 ml of water was calibrated by
dissolving a known amount of sodium chloride in the calorimeter.
The dissolution of sodium chloride in water can be represented by the
equation:
NaCl(s) + aq
Nat(aq) + Cl(aq)
AH+3.90 kJ mol-¹
The temperature of the calorimeter was initially 18.00 °C. A sample of 2.55 g of
sodium chloride was dissolved in the 100 mL and the temperature was
recorded as 16.30 °C.
The calibration factor of the calorimeter is
(express your answer to 2 decimal places)
0.42
kJ °C-1
Transcribed Image Text:A well insulated calorimeter containing 100 ml of water was calibrated by dissolving a known amount of sodium chloride in the calorimeter. The dissolution of sodium chloride in water can be represented by the equation: NaCl(s) + aq Nat(aq) + Cl(aq) AH+3.90 kJ mol-¹ The temperature of the calorimeter was initially 18.00 °C. A sample of 2.55 g of sodium chloride was dissolved in the 100 mL and the temperature was recorded as 16.30 °C. The calibration factor of the calorimeter is (express your answer to 2 decimal places) 0.42 kJ °C-1
Expert Solution
Step 1

Mass of NaCl used m = 2.55 g

Molar mass of NaCl = 1 x Na + 1 x Cl = 1 x 22.99 + 1 x 35.45 g/mol = 22.99 + 35.45 = 58.44 g/mol

Number of moles of NaCl used is mass of NaCl/molar mass of NaCl = 2.55 g/58.44 g/mol = 0.0436 moles

Heat of a solution, H = +3.90 kJ/mol

Energy of solution, E = H x number of moles of NaCl = +3.90 kJ/mol x 0.0436 moles = 0.17004 kJ 

 

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