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ycow [4]
3 years ago
13

A coffee-cup calorimeter initially contains 125 g water at 24.28C. Potassium bromide (10.5 g), also at 24.28C, is added to the w

ater, and after the KBr dissolves, the final temperature is 21.18C. Calculate the enthalpy change for dissolving the salt in J/g and kJ/mol. Assume that the specific heat capacity of the solution is 4.18 J/8C
Chemistry
1 answer:
iren2701 [21]3 years ago
6 0

Answer:

The solution is given below

Explanation:

Heat, q= mc∆T

q= 125g x 4.18 J/g∙°C x (21.18x- 24.28) °C

q=  -1619.75J

NEGATIVE SIGN INDICATES THAT HEAT IS ABSORBED.

Enthalpy Change, ∆H = 1619.75 7/ 10.5 g

                                     = 154.26 J/g

No. of moles of KBr = Mass of KBr/ Molecular Weight of KBr

                                =10.5g/119gmol-1

                                =0.088 mol

∆H= 1619.75 J/ 0.088 mol

      = 18.41 kJ/mol  

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Calculate the mole fraction of kbr (molar mass 119.00 g/mol) in a solution made by dissolving 0.30 g kbr in 0.400 l of H2O (d =
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The mole fraction of KBr in the solution is 0.0001

<h3>How to determine the mole of water</h3>

We'll begin by calculating the mass of the water. This can be obtained as follow:

  • Volume of water = 0.4 L = 0.4 × 1000 = 400 mL
  • Density of water = 1 g/mL
  • Mass of water =?

Density = mass / volume

1 = Mass of water / 400

Croiss multiply

Mass of water = 1 × 400

Mass of water = 400 g

Finally, we shall determine the mole of the water

  • Mass of water = 400 g
  • Molar mass of water = 18.02 g/mol
  • Mole of water = ?

Mole = mass / molar mass

Mole of water = 400 / 18.02

Mole of water = 22.2 moles

<h3>How to de terminethe mole of KBr</h3>
  • Mass of KBr = 0.3 g
  • Molar mass of KBr = 119 g/mol
  • Mole of KBr = ?

Mole = mass / molar mass

Mole of KBr = 0.3 / 119

Mole of KBr = 0.0025 mole

<h3>How to determine the mole fraction of KBr</h3>
  • Mole of KBr = 0.0025 mole
  • Mole of water = 22.2 moles
  • Total mole = 0.0025 + 22.2 = 22.2025 moles
  • Mole fraction of KBr =?

Mole fraction = mole / total mole

Mole fraction of KBr = 0.0025 / 22.2025

Mole fraction of KBr = 0.0001

Learn more about mole fraction:

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