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babymother [125]
1 year ago
5

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 =

1.00 g/ml; molar mass 18.02 g/mol).
Chemistry
1 answer:
julia-pushkina [17]1 year ago
6 0

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:

brainly.com/question/2769009

#SPJ1

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8 0
3 years ago
The enthalpy of solution (dissolving) of sodium hydroxide is given below. Determine the change in temperature of a coffee cup ca
geniusboy [140]

Answer:

The change in temperature of a coffee cup calorimeter is 8.87°C.

Explanation:

Volume of the water = V = 150 g

Density of the water , d =1.0 g/mL

Mass of the water = M

M=d\times V=1.00 g/mL\times 150 ml =150.0 g

Mass of solution = m = M = 150.0 g

NaOH(s)\rightarrow NaOH(aq),\Delta H =-44.51 kJ/mol

Moles of NaOH = \frac{5.00 g}{40 g/mol}=0.125 mol

Energy released when 0.125 moles of NaOH added in water = Q

Q=0.125 moles\times (-44.51 kJ/mol)=-5.5638 kJ=-5,563.8 J

1 kJ = 1000 J

Heat gained by water = Q' = -Q ( conservation of energy)

Q'= 5,563.8 J

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

Change in temperature of the solution = \Delta T

Q'=mc\times \Delta T

5,563.8 J=150.0 g\times 4.184 J/g^oC\times \Delta T

\Delta T=\frac{5,563.8 J}{150.0 g\times 4.184 J/g^oC}=8.87^oC

The change in temperature of a coffee cup calorimeter is 8.87°C.

7 0
3 years ago
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