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lbvjy [14]
3 years ago
6

What is the boiling point of a solution that contains 3 moles of KBr in 2000 g of water? (Kb = 0.512 C/m)

Chemistry
1 answer:
pochemuha3 years ago
4 0

Answer:

The answer to your question is   Tb = 100.768°C

Explanation:

Data

Tb = ?

moles of KBr = 3

mass of water = 2000 g

Kb = 0.512 C/m

Process

1.- Calculate the molality of the solution

molality = moles of solute/mass of solvent (kg)

-Substitution

molality = 3/2

-Result

molality = 1.5

2.- Calculate ΔT

    ΔTb = Kbm

-Substitution

    ΔTb = (0.512)(1.5)

-Result

    ΔTb = 0.768

3.- Calculate the boiling point, the boiliing point of pure water is 100°C

   ΔTb = Tb - Tb(solvent)

-Solve for Tb

   Tb = Tb(solvent) + ΔTb

-Substitution

   Tb = 100 + 0.768

-Result

  Tb = 100.768°C

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Answer:

3.10×10¯⁵ ft³.

Explanation:

The following data were obtained from the question:

Density (D) of lead = 11.4 g/cm³

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Density (D) = mass (m) / Volume (V)

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Finally, we shall convert 0.877 cm³ to ft³. This can be obtained as follow:

1 cm³ = 3.531×10¯⁵ ft³

Therefore,

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See the explanation below, please.

Explanation:

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