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Butoxors [25]
3 years ago
13

What is the molality of a solution in which 0.32 moles AlCl3 has been dissolved in 2,200 g water? What mass of water is needed t

o prepare a 1.20 molal solution using 0.60 mol propyleneglycol? What is the molality of a solution in which 0.145 mol CO2 is dissolved in 591 g water?
Chemistry
1 answer:
klemol [59]3 years ago
8 0

Answer:

1) The molality is 0.145 molal

2) We need 0.50 kg water

3) The molality is 0.245 molal

Explanation:

What is the molality of a solution in which 0.32 moles AlCl3 has been dissolved in 2,200 g water?

Molality = moles AlCl3 / mass water (in kg)

Molality = 0.32 moles / 2.2 kg water

Molality = 0.145 moles/kg = 0.145 molal

The molality is 0.145 molal

What mass of water is needed to prepare a 1.20 molal solution using 0.60 mol propyleneglycol?

Molality = moles propyleneglycol / mass water

1.20 = 0.60 / x kg

X = 0.60 / 1.2

X = 0.50 kg water

We need 0.50 kg water

What is the molality of a solution in which 0.145 mol CO2 is dissolved in 591 g water?

Molality = moles CO2 / mass water (in kg)

Molality = 0.145 moles / 0.591 kg

Molality = 0.245 molal

The molality is 0.245 molal

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Answer:

The bond dissociation energy to break 4 bonds in 1 mol of CH is 1644 kJ

Explanation:

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From the table one mole is C-H bond requires 411 kJ, that is 411 kJ/mol. Therefore, 4 C-H bonds would require 4 × 411 kJ = 1644 kJ

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                      %age Yield =   96 %

Explanation:

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Step 1: <u>Calculate moles of Ba(NO₃)₂:</u>

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Moles  =  75.1 g / 261.33 g/mol

Moles  =  0.2873 moles of Ba(NO₃)₂

Step 2: <u>Find out moles of BaSO₄ formed:</u>

According to balance chemical equation,

                  1 mole of Ba(NO₃)₂ produced  =  1 mole of BaSO₄

So,

        0.2873 moles of Ba(NO₃)₂ will produce  =  X moles of BaSO₄

Solving for X,

                      X =  0.2873 mol × 1 mol / 1 mol

                       X =  0.2873 moles of BaSO₄

Step 3: Calculate Theoretical Mass of BaSO₄:

Mass  =  Moles × M.Mass

Mass  =  0.2873 mol × 233.38 g/mol

Mass  = 67.07 g of BaSO₄

Step 4: <u>Calculate %age Yield as:</u>

                 Theoretical Yield  =  67.07 g

                  Actual Yield  =  64.4 g

                  %age Yield  =  <u>???</u>

Formula Used:

                   %age Yield  =  (Actual Yield ÷ Theoretical Yield) × 100

Putting Values,

                   %age Yield  =  (64.4 g ÷ 67.07 g) × 100

                   %age Yield =  96.01 % ≈ 96 %

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P1V1 = K         P2V2 = K

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