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lbvjy [14]
4 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:
pochemuha4 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:

Na₂CO₃.2H₂O

Explanation:

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Given values:

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To determine the integer x, we find the number of moles of the anhydrous Na₂CO₃ and that of the water of crystallisation:

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Molar mass of H₂O = [(1x2) + (16)] = 18gmol⁻¹

Number of moles of Na₂CO₃ = \frac{3.22}{106} = 0.03mole

Number of moles of H₂O =  \frac{1.09}{18} = 0.06mole

From the obtained number of moles:

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                           0.03                                    0.06

Simplest

Ratio                  0.03/0.03                         0.03/0.06

                                 1                                      2

Therefore, x = 2    

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